Gradient methods with retards and generalizations
GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS ?
A.FRIEDLANDER ?,J.M.MART ′
INEZ ?,B.MOLINA ?,AND M.RAYDAN ?SIAM J.N UMER.A NAL .c
1999Society for Industrial and Applied Mathematics Vol.36,No.1,pp.275–289
Abstract.A generalization of the steepest descent and other methods for solving a large scale
symmetric positive de?nitive system Ax =b is presented.Given a positive integer m ,the new iteration is given by x k +1=x k ?λ(x ν(k ))(Ax k ?b ),where λ(x ν(k ))is the steepest descent step at a previous iteration ν(k )∈{k,k ?1,...,max {0,k ?m }}.The global convergence to the solution of the problem is established under a more general framework,and numerical experiments are performed that suggest that some strategies for the choice of ν(k )give rise to e?cient methods for obtaining approximate solutions of the system.
Key words.gradient method,Barzilai–Borwein method,Rayleigh quotient,conjugate gradient method,symmetric successive overrelaxation (SSOR)preconditioning strategy
AMS subject classi?cations.65F10,49M07,65F15PII.S003614299427315X
1.Introduction.We are interested in the problem Minimize f (x )≡
12
x t
Ax ?b t x,(1.1)
where A is a real symmetric positive de?nite (SPD)n ×n matrix.The global solu-tion of (1.1)is the unique solution of Ax =b .The well-known steepest descent (or gradient)method is de?ned by the iteration
x k +1=x k ?λ(x k )g (x k ),
where λ(x )is the minimizer of ?(λ)≡f (x ?λg (x ))and g (x )=Ax ?b .Hence,λ(x )=
g (x )t g (x )
g (x )t Ag (x )
(1.2)
for all x ∈R n such that g (x )=0.This classical method is globally convergent and its storage requirements are minimal,but its rate of convergence is very low in critical cases (see [11]).
Barzilai and Borwein [3]proposed the iteration
x k +1=x k ?λ(x k ?1)g (x k )
for solving Ax =b ,where the ?rst steplength “λ(x ?1)”is arbitrary.Raydan [18]proved that the Barzilai–Borwein iteration is globally convergent and showed [17]that it is much more e?cient than the steepest descent method for solving large scale positive de?nite linear systems of equations.
?Received
by the editors August 22,1994;accepted for publication (in revised form)January 26,
1998;published electronically,January 5,1999.
https://www.wendangku.net/doc/0614065105.html,/journals/sinum/36-1/27315.html
?Department of Applied Mathematics,IMECC-UNICAMP,University of Campinas,CP 6065,13081-970Campinas SP,Brazil (friedlan@ime.unicamp.br,martinez@ime.unicamp.br).The ?rst and second authors were supported by FAPESP grant 90-3724-6,FINEP,and FAEP-UNICAMP.?Departamento de Computaci′o n,Facultad de Ciencias,Universidad Central de Venezuela,Ap.47002,Caracas 1041-A,Venezuela (bmolina@reacciun.ve,mraydan@reacciun.ve).The third author was supported by the Parallel and Distributed Computing Center at UCV.The fourth author was supported by BID-CONICIT project M-51940and FAEP-UNICAMP.
275
276 A.FRIEDLANDER,J.M.MART ′INEZ,B.MOLINA,AND M.RAYDAN
The gradient method with retards,introduced in this paper,is a generalization
of the steepest descent and the Barzilai–Borwein methods.Given a positive integer m ,the new iteration is x k +1=x k ?λ(x ν(k ))g (x k ),
(1.3)
where ν(k )is arbitrarily chosen in the set {k,k ?1,...,max {0,k ?m }}.A practical motivation for the introduction of this iteration is that it allows us to compute λ(x k )simultaneously in parallel with x k +1.Observe that,even in the e?cient conjugate gradient method,the computation of the steplength associated to the search direction necessarily precedes the computation of the new iterate.In this paper,we prove that the iteration (1.3)is globally convergent to the solution of the problem.From the practical point-of-view,we give numerical experiments that suggest that some strategies for the choice of ν(k )give rise to e?cient methods for obtaining approximate solutions of Ax =b .
In the convergence analysis we state the iteration (1.3)under a more general framework.Observe that,by (1.2),the recurrence (1.3)is a particular case of x k +1=x k ?
1αk
g k ,(1.4)
where g k =g (x k )and the scalar αk is the Rayleigh quotient of the matrix A at an n -dimensional vector,whose choice depends on the index k .For example,Barzilai and Borwein gave a second choice for the steplength that is covered by (1.4)but not by (1.3).This choice is αk =y t k ?1y k ?1s t k ?1y k ?1=s t k ?1A 2s k ?1
s t k ?1As k ?1
,
(1.5)
where s k ?1=x k ?x k ?1and y k ?1=g k ?g k ?1.In this case,αk is the Rayleigh
quotient of A at the vector √As k ?1.Our analysis corresponds to the iteration (1.4)for a family of choices of αk that include (1.3)and (1.5).
This paper is organized as follows.In section 2we establish the convergence results.The analysis that we present is an extension of the analysis presented in [18]for the particular choice of the Barzilai–Borwein method.The structure of the proofs will be the same as in [18].However,important di?erences come from the fact that now the scalars αk might be arbitrarily chosen from a ?nite collection of candidates.In section 3,we discuss this collection of possible practical choices,and we establish that some well-known methods can now be viewed as particular cases of this new scheme.We present some numerical experiments to demonstrate the potential of this new method.In section 4,we discuss preconditioning strategies.We present numerical results for elliptic partial di?erential equations to illustrate the behavior of this method when compared to the preconditioned conjugate gradient (PCG)method.Finally,in section 5we present concluding remarks.
2.Convergence analysis.Let x ?be the unique minimizer of f .Assume that {x k }is the sequence generated by (1.4)from a given vector x 0,where αk is an arbitrary
Rayleigh quotient,that is αk =u t k Au k /u t
k u k for some u k =0.We de?ne e k =x ??x k for all k .Using (1.4)and the fact that g k =Ax k ?b ,we have Ae k =αk s k
for all k.
(2.1)
GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS
277
Substituting s k =e k ?e k +1in (2.1)we obtain for any k e k +1=
1
αk
(αk I ?A )e k .(2.2)
Now for any initial error e 0,there exist constants d 01,d 02,...,d 0
n such that
e 0=
n i =1
d 0i v i ,
where {v 1,v 2,...,v n }are orthonormal eigenvectors of A associated with the eigen-values {σ1,σ2,...,σn }.Throughout this work we assume that
0<σmin =σ1≤σ2≤···≤σn =σmax .
Using (2.2)we obtain for any integer k ,e k +1=
n i =1
d k +1
i
v i ,(2.3)where d k +1
i
=
αk ?σi
αk
d k i
=k j =0
αj ?σi
αj d 0i .(2.4)
The convergence properties of the sequence {e k }depend on the behavior of each
one of the sequences {d k i },1≤i ≤n .In general,these sequences increase at some iterations.However,the following lemma shows that the sequence {d k 1}decreases Q-linearly to zero independently of the Rayleigh quotient of A that is being used,at every k ,to obtain the scalar αk .
Lemma 2.1.If the scalar αk in (1.4)is always chosen as the Rayleigh quotient of A at any arbitrary nonzero vector,then the sequence {d k 1}converges to zero Q-linearly with convergence factor ?c =1?(σmin /σmax ).
Proof.For any positive integer k ,
d k +1
1=
αk ?σmin αk d k 1.Since αk satis?es 0<σmin ≤αk ≤σmax
(2.5)
for all k ,we have
|d k +1
1
|=
1?σmin
αk
|d k 1|≤?c |d k 1|,where ?c =1?σmin
σmax
<1.We now describe the collection of possible choices of the scalar αk that allows us to obtain global convergence of the iteration (1.4).Given a positive integer m ,and q j ≥1,j =1,...,m a set of real numbers,we set
αk =
e t ν(k )A
ρ(k )
e ν(k )e t ν(k )A
(ρ(k )?1)e ν(k ),
(2.6)
278 A.FRIEDLANDER,J.M.MART ′INEZ,B.MOLINA,AND M.RAYDAN
where
ν(k )∈{k,k ?1,...,max {0,k ?m }}
and
ρ(k )∈{q 1,...,q m }
for all k =0,1,2,....
Notice that the scalar αk given by (2.6)is the Rayleigh quotient of A at the vector √A (ρ(k )?1)e ν(k ),and so it satis?es (2.5)for all k .Later,in section 3,we discuss some well-known methods and some new methods that can be embedded into the scheme given by (1.4)and (2.6).In the proof of our convergence theorem,we will use the following result.
Lemma 2.2.Let {e k }be the sequence of errors generated by the method given
by (1.4)and (2.6).If the sequences {d k 1},{d k 2},...,{d k
l },de?ned in (2.4)converge to zero for a ?xed integer l ,1≤l lim inf k →∞ |d k l +1|=0. Proof .Suppose,by way of contradiction,that there exists a constant ε>0such that (d k l +1) 2 min 1≤j ≤m σ(q j ?1) l +1≥εfor all k. (2.7) By (2.3),(2.6),and the orthonormality of the eigenvectors {v 1,v 2,...,v n },the Rayleigh quotient αk can be written as αk = n r =1(d ν(k )r )2σρ(k ) r n r =1(d ν(k )r )2σ(ρ(k )?1)r .(2.8)Since the sequences {d k 1},...,{d k l }all converge to zero,there exists ?k su?ciently large such that max 1≤j ≤m l r =1 (d k r )2σ(q j ?1)r ≤ε2for all k ≥?k.(2.9)By (2.8)and (2.9),we obtain for any k ≥?k +m ,( n r =l +1(d ν(k )r )2σ(ρ(k )?1) r )σl +1ε 2 +( n r =l +1(d ν(k )r )2σ(ρ(k )?1)r )≤αk ≤σmax .(2.10)Since n r =l +1 (d ν(k )r )2σ(ρ(k )?1) r ≥(d ν(k ) l +1)2σ(ρ(k )?1) l +1≥ε, then it follows from (2.10)that 2 3 σl +1≤αk ≤σmax for all k ≥?k +m, GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS 279 which implies the bound 1?σl +1αk ≤max 12,1?σl +1σmax for all k ≥?k +m.(2.11)Finally,using (2.11)and the ?rst part of (2.4),we obtain for all k ≥?k +m ,|d k +1l +1|= 1?σl +1αk |d k l +1|≤?c |d k l +1|,where ?c =max 1 2,1?σmin σmax <1. (2.12) Because this conclusion contradicts the hypothesis (2.7),we ?nd that the lemma is true. Theorem 2.1establishes the convergence of the method de?ned by (1.4)and (2.6)when applied to a quadratic function with a symmetric positive de?nite Hessian. Theorem 2.1.Let f (x )be a strictly convex quadratic function.Let {x k }be the sequence generated by (1.4)and (2.6)and x ?the unique minimizer of f .Then,either x j =x ?for some ?nite j ,or the sequence {x k }converges to x ?. Proof .We need only consider the case in which there is no ?nite integer j such that x j =x ?.Hence,it su?ces to prove that the sequence {e k }converges to zero.From (2.3)and the orthonormality of the eigenvectors we have e k 22 = n r =1 (d k r )2 . Therefore,the sequence of errors {e k }converges to zero if and only if each one of the sequences {d k i }for i =1,2,...,n converges to zero. Lemma 2.1shows that {d k 1}converges to zero.We prove that {d k p }converges to zero for 2≤p ≤n by induction on p .Therefore,we let p be any integer from this interval,and we assume that {d k 1},...,{d k p ?1}all tend to zero.Then for any given ε>0there exists ?k su?ciently large such that max 1≤j ≤m p ?1 r =1 (d k r )2σ(q j ?1) r <ε2 for all k ≥?k. (2.13) From (2.8)and (2.13),we obtain ( n r =p (d ν(k )r )2σ(ρ(k )?1) r )σp ε 2+( n r =p (d ν(k )r )2σ(ρ(k )?1)r )≤αk ≤σmax ,(2.14) for all integers k ≥?k +m .Moreover,by Lemma 2.2,there exists k p ≥?k +m such that (d k p p )2 max 1≤j ≤m σ(q j ?1)p <ε. 280 A.FRIEDLANDER,J.M.MART ′INEZ,B.MOLINA,AND M.RAYDAN Now,consider any integer k 0>k p such that max j (d k 0?1p )2σ(q j ?1) p <εand max j (d k 0p )2σ(q j ?1)p ≥ε.Clearly,for k 0≤k ≤j 0?1, max 1≤j ≤m n r =p (d k r )2σ(q j ?1) r ≥max 1≤j ≤m (d k p )2σ(q j ?1)p ≥ε,(2.15) where j 0is the ?rst integer greater than k 0for which max j (d j 0p )2 σ(q j ?1) p <ε.Then,by (2.14)and (2.15),and using an argument similar to the one used in the proof of Lemma 2.2,we have 2 3 σp ≤αk ≤σmax for k 0≤k ≤j 0?1. (2.16) Thus,using (2.16)and the ?rst part of (2.4),we obtain |d k +2p |≤?c |d k +1p | for k 0≤k ≤j 0?1, where ?c is the constant (2.12),which satis?es ?c <1.Finally,using the bound |d k 0+1p |≤ σmax ?σmin σmin 2|d k 0?1p |,which is implied by expression (2.5)and the ?rst part of (2.4),we conclude that (d k p )2≤ σmax ?σmin σmin 4(d k 0?1p )2≤ σmax ?σmin σmin 4 ε(max j σ(q j ?1)p ) for all k 0+1≤k ≤j 0+1.Further,(2.4)provides the inequality (d k 0p )2 ≤((σmax ? σmin )/σmin )2(d k 0?1p )2.It follows from the conditions on k 0,k p ,and j 0that (d k p )2 is bounded above by a constant multiple of εfor all k ≥k p .Hence,since ε>0can be chosen to be arbitrarily small,we deduce lim k →∞|d k p |=0as required,which completes the proof. It is worth mentioning that a straightforward extension of Theorem 2.3can be made for the (not necessarily strictly)convex case.This extension has been done in [9]for the Barzilai–Borwein method. 3.Classical and new choices of αk .If ρ(k )=3for all k ,we see,using (2.1)that formula (2.6)takes the form αk = s t ν(k )As ν(k )s t ν(k )s ν(k ) = g t ν(k )Ag ν(k )g t ν(k )g ν(k ) . (3.1) Therefore,in this case the method de?ned by (1.4)and (2.6)is the gradient method with retards introduced in section 1.On the other hand,the second method of Barzilai and Borwein,given by (1.5),corresponds to ρ(k )≡4and ν(k )≡k ?1.Consequently,the methods de?ned by ρ(k )≡4are generalizations of the second Barzilai–Borwein method. We can de?ne the following gradient methods with retards,with the same cost of the steepest descent method,both in terms of storage and work per iteration.We de?ne ?rst k =max {0,k ?m }.ν(k )=random integer between k and k ; (3.2) GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS281 ν(k)=k ifν(k?1) (3.3) ν(k)=k; (3.4) ν(k)=argmax{λ(x k),...,λ(x k )}; (3.5) ν(k)=argmin{λ(x k),...,λ(x k )}; (3.6) ν(k)=k if k is even,ν(k)=k if k is odd. (3.7) ν(k)=random integer between k and k?1. (3.8) From now on,we consider only gradient methods with retards as de?ned by(1.3). The following result should help us to decide which strategies,among those de?ned above,could be e?cient. Theorem3.1.Let{s k}be the sequence generated by(1.4)and(3.1).Assume that the sequence{s k/ s k }is convergent,that is,there exists s∈R n, s =1,such that lim k→∞ s k s k =s. (3.9) Then lim k→∞αk=s t As, (3.10) s is an eigenvector of A with eigenvalue s t As and the convergence of{x k}is Q-superlinear. Proof.By(3.1)we have that αk= s t ν(k) sν(k) A sν(k) sν(k) , (3.11) whereν(k)≥k?m.By(3.9), lim k→∞ sν(k) sν(k) =s, so(3.10)follows taking limits on both sides of(3.11). Now,combining(2.1)and(2.2)it follows that s k+1=? 1 αk+1 (A?αk I)s k. Therefore, s k+1 s k+1 =? (A?αk I)s k/ s k (A?αk I)s k/ s k . (3.12) 282 A.FRIEDLANDER,J.M.MART′INEZ,B.MOLINA,AND M.RAYDAN De?neα=s t As and c= (A?αI)s =lim k→∞ (A?αk I)s k/ s k . Suppose for a moment that c=0.Taking limits on both sides of(3.12)we obtain s=?1 c (A?αI)s. Therefore, (A?(α?c)I)s=0. This means that s is an eigenvector of A,with eigenvalueα?c.Therefore,α= s t As=α?c and,thus,c=0.So,c= (A?αI)s =0and,consequently,s is an eigenvector of A with eigenvalueα,as we wanted to prove.Hence, lim k→∞(A?αk I)s k s k =(A?αI)s=0. (3.13) But(3.13)is the Dennis–Mor′e su?cient condition(see[5])for superlinear convergence of the quasi-Newton iteration x k+1=x k?B?1k g(x k).In this case,B k=αk I.Thus, the sequence{x k}converges Q-superlinearly,as we wanted to prove. 3.1.Numerical experiments.In addition to the gradient methods with re-tards,we used the classical conjugate gradient method for solving our test problems (see[11],[14]).The conjugate gradient method uses one n-dimensional vector more than the gradient methods with retards,and requires a little more computational work per iteration.However,as it is well known,the conjugate gradient method has ex-cellent convergence properties,including?nite termination in n iterations and is the method of choice for solving large-scale symmetric positive de?nite linear systems. See,for example[2]. Iterative linear methods are very important in the context of iterative resolution of large-scale nonlinear systems of equations.In this case the“Newtonian linear sys-tem”must be solved approximately at each iteration of the nonlinear solver.See for instance,[4],[6],[7],[15],and[20].A very high precision is neither necessary nor recommendable in the resolution of these linear systems and the danger of“oversolv-ing”these subproblems may lead to unacceptable computer times in the resolution of the main problems.Thus,mild stopping criteria are used.The best-known stopping criterion(see[4])imposes that the residual at a given iterate should be a fraction of the residual at the initial iteration.In our notation,this turns out to be Ax k?b ≤θ Ax0?b . (3.14) Other situations where mild stopping criteria like(3.14)are used,when the main problem is more general than the linear system,may be found in[8],[9],and[13]. In our tests we used the stopping criterion(3.14)withθ=10?1,10?2,10?3, and10?4.All our tests were done on an HP Apollo135/125workstation in double precision Fortran.The methods tested are listed below. “0”Conjugate gradient method; “1”Steepest descent method; GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS283 Table3.1 Average of(iterations)/(?ops×10?3)using?ve initial points. Methodθ=10?1θ=10?2θ=10?3θ=10?4 03/189/5428/16892/552 14/2016/8083/415414/2070 23/159/4532/16072/360 33/1515/7533/165110/550 43/1115/5138/128140/476 53/1116/7042/200100/490 64/2018/9086/430503/2515 73/1511/5535/175210/1050 83/1513/6527/13576/380 93/1512/6040/20082/410 “2”Barzilai–Borwein method(method based on(3.4)with m=1); “3”Random choice ofν(k)(3.2); “4”Cyclic choice ofν(k)(3.3); “5”Maximum retard(method based on(3.4)with m=5); “6”Maximumλ(method based on(3.5); “7”Minimumλ(method based on(3.6); “8”Maximum-minimum retard(method based on(3.7); “9”Random choice ofν(k),excluding steepest descent(3.8). We now describe two di?erent test problems and the corresponding numerical results. Two point boundary value problems:Consider the matrix A≡(a ij)given by a ii=2/h2,a i,i?1=?1/h2if i=1,a i,i+1=?1/h2if i=n, for i=1,...,n,where h=11/n and n=1000.Linear systems of this kind appear frequently in the numerical solution of two point boundary value problems. We generated a random solution x?with components between?10and10and we computed b=Ax?.We used?ve initial points x0,originated by di?erent seeds of the random number generator.In Table3.1we report the average number of iterations and the average computational work required by each method using the?ve initial points,and m=5,to achieve the stopping criterion with the . ∞norm.The computational work is reported in?oating point operations(?ops)divided by103. The preliminary numerical experiments reported in Table3.1seem to indicate that,among the di?erent strategies for the gradient method with retards,the most e?cient are“2”(Barzilai–Borwein),“3”(random),“4”(cyclic),“5”(maximum re-tard),“8”(maximum-minimum retard),and“9”(random excluding steepest descent). On the other hand,the steepest descent method,together with the strategies based on (3.5)and(3.6),are clearly less e?cient.It is interesting to observe that the successful gradient methods with retards appear to use about the same number of iterations as the conjugate gradient method for obtaining the precision(3.14),for the tested values ofθ.This can be a very useful feature of these methods since,as we mentioned before, they can be implemented using less computer time and storage than the conjugate gradient method.(For this particular problem,the computer time of one iteration of the gradient methods with retards is about0.8times the computer time of one itera-tion of the conjugate gradient method).Moreover,for some of the gradient methods 284 A.FRIEDLANDER,J.M.MART ′INEZ,B.MOLINA,AND M.RAYDAN with retards,a steplength does not have to be computed at every iteration,and that could represent a considerable reduction in the computational work required during the process.For example,the cyclic choice of ν(k )(method 4)only needs to compute one steplength every m +1iterations. Random problems:We take A =QDQ t ,where Q =(I ?2w 3w T 3 )(I ?2w 2w T 2)(I ?2w 1w T 1),w 1,w 2,and w 3are unitary random vectors,D =Diag (σ1,...,σn )is a diagonal matrix where σ1=1,σn =cond ,and σj is randomly generated between 1and cond for j =2,...,n ?1.The entries of the right-hand-side b are randomly generated between ?10and 10.The initial point is the null vector of R n and,in these tests,we used n =5000and we allowed a maximum of 1,000iterations. In Table 3.2we report the number of iterations used to obtain Ax k ?b 2≤θ b 2by the CG method and some of the most e?cient gradient methods with retards.We observe that the di?erence between the number of iterations used by di?erent methods is not great.For large condition numbers,the performance of gradient methods with retards deteriorates.However,it is very interesting to observe that,for cond =107all the gradient methods with retards achieved the precision 0.1in about 40iterations,while the method of conjugate gradients used more than 400iterations for the same purpose.This behavior favors the application of gradient methods with retards in the inexact-Newton context,as we mentioned before. Since,in principle,all the gradient methods with retards,including steepest de-scent,satisfy the same convergence theorem,a more careful analysis is necessary to understand the numerical experiments.The following remarks are based on the available theory and tend to explain the numerical behavior observed. (a)The hypothesis of Theorem 3.1(superlinear convergence)is the convergence of the normalized gradients.This hypothesis cannot be veri?ed if we choose the steepest descent method or the strategy based on (3.5),since,in both cases,the angle between successive gradients is at least 90degrees. (b)The “ideal”behavior of an iteration based on (1.4)could be obtained choosing {α0,α1,α2,...}as di?erent eigenvalues of A .By (2.2),if (say)αk =σi ,we have that d k +j i =0for all j ≥1.As a result,this ideal method should ?nd the solution in a number of iterations equal to the number of di?erent eigenvalues of A .It is interesting to observe that this property is independent of the choice of the initial point.This is precisely the main termination result of the conjugate gradient method.An e?cient gradient method with retards should approximate,in some sense,the ideal iteration,and numerical results suggest that the e?cient strategies are successful in doing that.In particular,Glunt,Hayden,and Raydan [10]established a relationship with the shifted power method that adds understanding to the practical behavior of strategy “2”(Barzilai–Borwein). 4.Preconditioned version.We present the preconditioned version of the gra-dient method with retards (PGMR),and compare its performance with the PCG method on some large and sparse SPD linear systems arising from the discretization of elliptic PDE problems.Our presentation follows the recent work by Molina and Raydan [16]for the preconditioned Barzilai–Borwein method. For practical purposes,we only consider the case ρ(k )=3for all k .In that case,one auxiliary linear system of equations needs to be solved per iteration,and the PGMR can be written as (see [16]): GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS 285 Table 3.2 Iterations required by CG,Barzilai–Borwein “2”,cyclic “4”,maximum retard “5”,and max-min retard “8”,for random problems. cond θCG 2458102 10?191917201710?2223027342610?3335048534710?44562576657103 10?1252123302210?2636670787010?310212412314012110?4132166160185162104 10?1643830353210?218515916919017310?326735834638934210?434061************* 10?12354133373310?233838640667039810?3412>1000>1000>1000>100010?4447106 10?13424334423510?2416>1000 >1000 >1000 >1000 10?345310?4485107 10?14174334433710?2453>1000>1000>1000>1000 10?348510?4 519 Algorithm 4.1.:PGMR Given x 0∈R n ,α0a nonzero real number, m a positive integer and C an SPD matrix of order n .Set g 0=Ax 0?b . For k =0,1,...(until convergence)do Choose ν(k )∈{k,k ?1,...,max {0,k ?m }}Solve Ch k =g k ,for h k Set p k =Ah k Set g k +1=g k ?1?αν(k ) p k Set x k +1=x k ?1?αν(k ) h k Set ?αk +1 =h t k p k g t k h k End do Notice that every iteration of the PGMR algorithm requires two inner products,two scalar-vector multiplications,two vector additions,one matrix-vector multiplica-tion,and the solution of a linear system of equations with the preconditioning matrix C .Since C is an SPD matrix,it follows from a straightforward application of Theorem 2.3that the sequence {x k }converges to x ?. Consider now the elliptic partial di?erential equation ? ?2u (x,y )?x 2??2u (x,y ) ?y 2 +?γu (x,y )=f 1(x,y ),(4.1) 286 A.FRIEDLANDER,J.M.MART′INEZ,B.MOLINA,AND M.RAYDAN Table4.1 (iterations/?ops×10?6)whenγ=0and n=25×104. Methodθ=10?1θ=10?4θ=10?8 PCG55/20789/334128/480 BarBor60/210100/350157/549 cyclic(m=3)70/218101/315170/530 cyclic(m=4)78/241115/356160/496 maximum(m=3)59/20698/343162/567 maximum(m=4)62/217100/349160/560 max-min(m=3)66/230108/377145/507 max-min(m=4)68/237120/438178/621 where?γ≥0and the function f1(x,y)are given.We solve(4.1)on the unit square 0≤x≤1,0≤y≤1,with homogeneous Dirichlet boundary conditions,i.e.,we wish to obtain a function u(x,y)that is continuous on the unit square,satis?es(4.1)in the interior of the unit square,and equals zero on the boundary. To obtain the numerical solution of this problem,we discretize(4.1)using the central?nite di?erence scheme of?ve points in a uniform grid p×p with a step h=1/(p+1)and the natural ordering,producing a system of linear equations Ax=b of order n=p2.The matrix A is SPD and has block tridiagonal structure. All the diagonal blocks of A are square,symmetric,and tridiagonal matrices of order p,and there are exactly p blocks in the diagonal of A.The diagonal elements of A are(4+γ),whereγ=h2?γ.For more details see[16]. The i th position of the vector b is equal to(f1)i,where(f1)i is the evaluation of f1in the i th node of the grid in the natural ordering.Finally,note that forγ=0our model problem reduces to the classical Poisson equation for which A is ill-conditioned. Whenγincreases,the condition number of A is gradually reduced. We ran an implementation of Algorithms PGMR and PCG,to solve problem (4.1),for di?erent values of the parameterγand di?erent step sizes h.We report results with the e?cient PCG version described in Golub and Van Loan[11],and the well-known symmetric successive overrelaxation(SSOR)preconditioning strategy for both algorithms(see,for instance,[1]).The parameterωassociated with the SSOR technique was chosen as in[16],for both methods.We also ran some experiments with the modi?ed incomplete Cholesky preconditioning strategy,introduced by Gustafsson [12].However,the results were quite similar to the ones obtained with the SSOR strategy,and we do not report them. We start the process in both algorithms at x0=(0,0,...0)t and we setα0=1in Algorithm PGMR.We stop the process,in both algorithms whenever g k 2≤θ g0 2 for di?erent values ofθ. In our?rst experiment we?xγ=0and p=500,i.e.,n=25×104.Table4.1shows the number of iterations and the computational work required by the PCG method and the PGMR with the following strategies(see section3):Barzilai–Borwein(Bar-Bor),cyclic retard(cyclic),maximum retard(maximum),and maximum-minimum retard(max-min).The computational work is reported in number of?ops divided by106.We can see that the di?erent strategies in the PGMR family are competitive with the PCG method.The PCG method shows a moderate improvement over the PGMR family in number of iterations.Nevertheless,when low accuracy is required, the winner in computational work by a slight di?erence is a member of the PGMR family.In particular,the case m=3seems to perform better than m=4in the three strategies that involve the parameter m. GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS287 Table4.2 (iterations/?ops×10?6)whenγ=0.1,m=3and n=25×104. Methodθ=10?1θ=10?4θ=10?8 PCG8/3017/6430/113 BarBor8/2819/6632/112 cyclic9/2818/5632/99 maximum9/3119/6632/112 max-min8/2818/6331/108 Fig.1.PCG and cyclic for n=25×104andθ=10?8. In our second experiment we study the e?ect of reducing the condition number of A by increasingγ.Table4.2shows the number of iterations and the computational work required by the PCG method and the PGMR when n=25×104,γ=0.1,and m=3.We observe that when the condition number of A is reduced,the PCG method and the PGMR tend to require the same number of iterations,and this number of it-erations decreases.As a consequence,some strategies in the PGMR show a moderate but consistent improvement over the PCG method in number of?ops.In particular, Figure1shows the number of?ops required by PCG and cyclic retard(m=3)for n=25×104,θ=10?8,and di?erent values ofγ. 5.Concluding remarks.We have introduced the gradient method with retards to solve symmetric and positive de?nite linear systems of equations.This method is a generalization of the steepest descent and the Barzilai and Borwein methods,and covers a wide range of possible strategies to choose steplengths that are de?ned by Rayleigh quotients. Based on our numerical results,we conclude that some of the new strategies are very e?cient and,in general,compete favorably with the classical conjugate gradient (CG)method in storage requirements and computational work,when low precision is required.In particular,we have no theoretical explanation for the outstanding behavior of the GMR for ill-conditioned problems whenθ=10?1(see Table3.2). However,if the coe?cient matrix is very ill-conditioned and high precision is required then the CG method is clearly a better option. 288 A.FRIEDLANDER,J.M.MART′INEZ,B.MOLINA,AND M.RAYDAN We also present a preconditioned version of the GMR and test it on large and sparse linear systems arising from the discretization of elliptic PDE problems.In this case,some of the PGMR strategies show a moderate but consistent improvement over the PCG method.In fact,the GMR is a gradient method for the minimization of quadratic functions.Therefore,as any gradient method for unconstrained minimiza-tion,its performance improves if the coe?cient(Hessian)matrix is conditioned to approximate the identity,in which case it behaves more like the Newton method. Finally,the possibility of computing the steplengths and the points in di?erent processors o?ers an additional advantage in terms of overall computer time.For example,when we consider the extension to minimize nonquadratic functions(see Raydan[19]),the steplengths can be expensive to compute,and so,it would be in-teresting to have the chance of using“old ones”while waiting for the computation of“new ones.”Furthermore,even in the sequential case,the possibility of repeating steplengths can be very time-saving.For instance,a gradient method that repeats the previous steplength whenever a nonmonotone criterion is satis?ed should be de-veloped,and deserves careful experimentation. Acknowledgments.We are grateful to the associate editor,Prof.P.L.Toint, and the referees for their constructive comments and suggestions. REFERENCES [1]O.Axelsson,A survey of preconditioning iterative methods for linear systems of algebraic equations,BIT,25(1985),pp.166–187. [2]R.Barrett,M.Berry,T.F.Chan,J.Demmel,J.Donato,J.Dongarra,V.Eijkhout, R.Pozo,Ch.Romine,and H.van der Vorst,Templates for the Solution of Linear Systems:Building Blocks for Iterative Methods,SIAM,Philadelphia,PA,1994. [3]J.Barzilai and J.M.Borwein,Two point step size gradient methods,IMA J.Numer.Anal., 8(1988),pp.141–148. [4]R.S.Dembo,S.C.Eisenstat,and T.Steihaug,Inexact Newton methods,SIAM J.Numer. Anal.,14(1982),pp.400–408. [5]J.E.Dennis and J.J.Mor′e,A characterization of superlinear convergence and its application to quasi-Newton methods,https://www.wendangku.net/doc/0614065105.html,p.,28(1974),pp.549–560. [6]P.Deuflhard,Global inexact Newton methods for very large scale nonlinear problems,Impact Comput.Sci.Engrg.,3(1991),pp.366–393. [7]S.C.Eisenstat and H.F.Walker,Globally convergent inexact Newton methods,SIAM J. Optim.,4(1994),pp.393–422. [8] A.Friedlander and J.M.Mart′?nez,On the maximization of a concave quadratic function with box constraints,SIAM J.Optim.,4(1994),pp.177–192. [9] A.Friedlander,J.M.Mart′?nez,and M.Raydan,A new method for large-scale box con- strained quadratic minimization problems,Optim.Methods Softw.,5(1995),pp.57–74. [10]W.Glunt,T.L.Hayden,and M.Raydan,Molecular conformations from distance matrices, https://www.wendangku.net/doc/0614065105.html,put.Chem.,14(1993),pp.114–120. [11]G.H.Golub and Ch.F.Van Loan,Matrix Computations,The Johns Hopkins University Press,Baltimore,London,1989. [12]I.Gustafsson,A class of?rst order factorization methods,BIT,18(1978),pp.142–156. [13]G.T.Herman,Image Reconstruction from Projections:The Fundamentals of Computerized Tomography,Academic Press,New York,1980. [14]M.R.Hestenes and E.Stiefel,Methods of conjugate gradients for solving linear systems, J.Research National Bureau of Standards,B49(1952),pp.409–436. [15]J.M.Mart′?nez,A theory of secant preconditioners,https://www.wendangku.net/doc/0614065105.html,p.,60(1993),pp.681–698. [16] B.Molina and M.Raydan,Preconditioned Barzilai–Borwein method for the numerical solu- tion of partial di?erential equations,Numer.Algorithms,13(1996),pp.45–60. [17]M.Raydan,Convergence Properties of the Barzilai and Borwein Gradient Method,Ph.D. thesis,Dept.of Mathematical Sciences,Rice University,Houston,TX,1991. [18]M.Raydan,On the Barzilai and Borwein choice of steplength for the gradient method,IMA J.Numer.Anal.,13(1993),pp.321–326. GRADIENT METHOD WITH RETARDS AND GENERALIZATIONS289 [19]M.Raydan,The Barzilai and Borwein gradient method for the large scale unconstrained minimization problem,SIAM J.Optim.,7(1997),pp.26–33. [20]T.J.Ypma,Local convergence of inexact Newton methods,SIAM J.Numer.Anal.,21(1984), pp.583–590. 英语介词用法大全 TTA standardization office【TTA 5AB- TTAK 08- TTA 2C】 介词（The Preposition）又叫做前置词，通常置于名词之前。它是一种虚词，不需要重读，在句中不单独作任何句子成分，只表示其后的名词或相当于名词的词语与其他句子成分的关系。中国学生在使用英语进行书面或口头表达时，往往会出现遗漏介词或误用介词的错误，因此各类考试语法的结构部分均有这方面的测试内容。 1. 介词的种类 英语中最常用的介词，按照不同的分类标准可分为以下几类： (1). 简单介词、复合介词和短语介词 ①.简单介词是指单一介词。如： at , in ,of ,by , about , for, from , except , since, near, with 等。②. 复合介词是指由两个简单介词组成的介词。如： Inside, outside , onto, into , throughout, without , as to as for , unpon, except for 等。 ③. 短语介词是指由短语构成的介词。如： In front of , by means o f, on behalf of, in spite of , by way of , in favor of , in regard to 等。 (2). 按词义分类 {1} 表地点（包括动向）的介词。如： About ,above, across, after, along , among, around , at, before, behind, below, beneath, beside, between , beyond ,by, down, from, in, into , near, off, on, over, through, throught, to, towards,, under, up, unpon, with, within , without 等。 {2} 表时间的介词。如： About, after, around , as , at, before , behind , between , by, during, for, from, in, into, of, on, over, past, since, through, throughout, till(until) , to, towards , within 等。 {3} 表除去的介词。如： beside , but, except等。 {4} 表比较的介词。如： As, like, above, over等。 {5} 表反对的介词。如： againt ,with 等。 {6} 表原因、目的的介词。如： for, with, from 等。 {7} 表结果的介词。如： to, with , without 等。 {8} 表手段、方式的介词。如： by, in ,with 等。 {9} 表所属的介词。如： of , with 等。 {10} 表条件的介词。如： 介词用法知多少 介词是英语中最活跃的词类之一。同一个汉语词汇在英语中可译成不同的英语介词。例如汉语中的“用”可译成：（1）用英语（in English）；（2）用小刀（with a knife）；（3）用手工（by hand）；（4）用墨水（in ink）等。所以，千万不要以为记住介词的一两种意思就掌握了这个介词的用法，其实介词的用法非常广泛，搭配能力很强，越是常用的介词，其含义越多。下面就简单介绍几组近义介词的用法及其搭配方法。 一. in, to, on和off在方位名词前的区别 1. in表示A地在B地范围之内。如： Taiwan is in the southeast of China. 2. to表示A地在B地范围之外，即二者之间有距离间隔。如： Japan lies to the east of China. 3. on表示A地与B地接壤、毗邻。如： North Korea is on the east of China. 4. off表示“离……一些距离或离……不远的海上”。如： They arrived at a house off the main road. New Zealand lies off the eastern coast of Australia. 二. at, in, on, by和through在表示时间上的区别 1. at指时间表示： （1）时间的一点、时刻等。如： They came home at sunrise （at noon, at midnight, at ten o’clock, at daybreak, at dawn）. （2）较短暂的一段时间。可指某个节日或被认为是一年中标志大事的日子。如： He went home at Christmas （at New Year, at the Spring Festival, at night）. 2. in指时间表示： （1）在某个较长的时间（如世纪、朝代、年、月、季节以及泛指的上午、下午或傍晚等）内。如： in 2004, in March, in spring, in the morning, in the evening, etc （2）在一段时间之后。一般情况下，用于将来时，谓语动词为瞬间动词，意为“在……以后”。如： He will arrive in two hours. 谓语动词为延续性动词时，in意为“在……以内”。如： These products will be produced in a month. 注意：after用于将来时间也指一段时间之后，但其后的时间是“一点”，而不是“一段”。如： He will arrive after two o’clock. 3. on指时间表示： （1）具体的时日和一个特定的时间，如某日、某节日、星期几等。如： On Christmas Day（On May 4th）, there will be a celebration. （2）在某个特定的早晨、下午或晚上。如： He arrived at 10 o’clock on the night of the 5th. （3）准时，按时。如： If the train should be on time, I should reach home before dark. 4. by指时间表示： （1）不迟于，在（某时）前。如： WITH/BY/IN的用法区别："The mountain is covered with/in/by snow". The meanings are so similar that the three can be used almost interchangeably, but some subtle nuances may apply. When referring to a substance(物体) that sticks to another, use in or with, but not by: ?The actress was covered in blood, or ?The actress was covered with blood, but not ?The actress was covered by blood. When referring something that physically protects something else, use with or by, but not in: ?The field(田地) was covered with a tarp(油布), or ?The field was covered by a tarp, but not ?The field was covered in a tarp. Use covered with to indicate an unusual amount of something on top of something else; use covered by to connote a covering so dense that the object being covered is completely obscured from view: ?The mountain was covered with fog. ?The mountain was covered by fog. Another example: ?Our grass(草坪) was covered with butterflies. ?Our grass was covered by butterflies. Somehow, the latter (covered by) paints a picture where the butterflies are so close together that I can hardly see the grass at all, but in the former (covered with), I picture a lot of butterflies, just not necessarily so many that I can't see the grass. When talking about metaphorical coverage, use covered by: ?The roof damage was covered by insurance(上保险), but not ?The roof damage was covered with insurance, or ?The roof damage was covered in insurance. Another example: ?The city council meeting was covered by the news station, but not ?The city council meeting was covered with the news station, or ?The city council meeting was covered in the news station. compare to 和compare with 的区别是什么 Compare to 是“把……比作”的意思。例如： We compare him to a little tiger. 我们把他比作小老虎。 The last days before liberation are often compared to the darkness before the dawn. 将要解放的那些日子常常被比作黎明前的黑暗。 Compare ... with 是“把……和……比较”的意思。例如： We must compare the present with the past. 我们要把现在和过去比较一下。 We compared the translation with the original. 我们把译文和原文比较了下。 从上面比较可以看出，compare with 侧重一个仔细的比较过程。有时，两者都可以互相代替。例如： He compared London to (with) Paris. 他把伦敦比作巴黎。 London is large, compared to (with) Paris. 同巴黎比较而言，伦敦大些。 在表示“比不上”、“不能比”的意思时，用compare with 和compare to 都可以。例如： My spoken English can't be compared with yours. 我的口语比不上你的。 The pen is not compared to that one. 这笔比不上那支。 1、c ompare…to…意为“把…比作”，即把两件事物相比较的同时，发现某些方面相似的地方。这两件被比较的事物 或人在本质方面往往是截然不同的事物。如： He compared the girl to the moon in the poem. 他在诗中把那姑娘比作月亮。 2、compare…with…“与…相比，把两件事情相比较，从中找出异同”，这两件事又往往是同类的， 如：I'm afraid my English compares poorly with hers. 恐怕我的英语同她的英语相比要差得多。 compare to和compare with有何区别,当说打比方时和做比较是分别用哪个？ compare…to…比喻.例如: The poets often compare life to a river. 诗人们经常把生活比喻成长河． compare…with…相比．例如： My English can't compare with his. 我的英文水平不如他． with的用法大全----四级专项训练with结构是许多英语复合结构中最常用的一种。学好它对学好复合宾语结构、不定式复合结构、动名词复合结构和独立主格结构均能起很重要的作用。本文就此的构成、特点及用法等作一较全面阐述，以帮助同学们掌握这一重要的语法知识。 一、 with结构的构成 它是由介词with或without+复合结构构成，复合结构作介词with或without的复合宾语，复合宾语中第一部分宾语由名词或代词充当，第二部分补足语由形容词、副词、介词短语、动词不定式或分词充当，分词可以是现在分词，也可以是过去分词。With结构构成方式如下： 1. with或without-名词/代词+形容词; 2. with或without-名词/代词+副词; 3. with或without-名词/代词+介词短语; 4. with或without-名词/代词+动词不定式; 5. with或without-名词/代词+分词。 下面分别举例： 1、 She came into the room，with her nose red because of cold.(with+名词+形容词，作伴随状语) 2、 With the meal over ， we all went home.(with+名词+副词，作时间状语) 3、The master was walking up and down with the ruler under his arm。(with+名词+介词短语，作伴随状语。) The teacher entered the classroom with a book in his hand. 4、He lay in the dark empty house，with not a man ，woman or child to say he was kind to me.(with+名词+不定式，作伴随状语) He could not finish it without me to help him.(without+代词 +不定式，作条件状语) 5、She fell asleep with the light burning.(with+名词+现在分词，作伴随状语) 6、Without anything left in the cupboard， she went out to get something to eat.(without+代词+过去分词，作为原因状语) 二、with结构的用法 在句子中with结构多数充当状语，表示行为方式，伴随情况、时间、原因或条件(详见上述例句)。 1.I don`t like a ___(rain) day. 2.Please tell Tom ___(call)me back. 3.____ is the weather like? It`s snowy. How. What. Where. Why. 4.____ ia it going? It`s great. How. What. Where. When. 5.Is he____? No, he’s ___ in the river. A. swims,fishing B. swimming, running C. swim,walking D. running, swimming 6.Hello!is that Linda speaking? Yes.______ Who’s that? Who’s it? Who are you? I’m Linda. 7.He often ____(watch)TV in the evening. But now he _____(read)a book. 8.My uncle is a ___(cook). 9.It’s a nice ____. A weather photo salad rice 10.After supper Mary often ___ along the river. Take a walk takes walk taking a walk takes a walk 11.How’s it going? _____. The weather is so cold. Preety good. Great. Terrible. Not bad. 12.Beijing ___ (have)a long history. 13.I am having a great time ____(visit)my grandmother. 14.We feel very ___ after the ___ (relax) vacation. 15.We are ___ vacation in Dalian. A in B on C at Dof 16.What ____ when it’s sunny? Aare you doing B do you do C does you D is he do 17.Tom like to go ___(skate) 18.Kate is very tired, so she needs ____(relax) 19.Would you like to ____ a photofor us? Sure. Agive B make C take Dget 20.Could you help me put up the maps on the wall?_____ A.No problem B. I hope so C. That’s all right D.That’s a good idea. 21.China plan to let Jim ____the Xisha Islands> Visit visits visiting visited 22.Where is Grace? She ____ IN THE yard. Reads read is reading was reading 23.Britain is ___ European country and Singapore is ___ Asian country A,a,a a, an the , an a ,the 1.I don`t like a ___(rain) day. 2.Please tell Tom ___(call)me back. 3.____ is the weather like? It`s snowy. How. What. Where. Why. 4.____ ia it going? It`s great. How. What. Where. When. 5.Is he____? No, he’s ___ in the river. A. swims,fishing B. swimming, running C. swim,walking D. running, swimming 6.Hello!is that Linda speaking? Yes.______ Who’s that? Who’s it? Who are you? I’m Linda. 7.He often ____(watch)TV in the evening. But now he _____(read)a book. 8.My uncle is a ___(cook). 9.It’s a nice ____. A weather photo salad rice 10.After supper Mary often ___ along the river. Take a walk takes walk taking a walk takes a walk 11.How’s it going? _____. The weather is so cold. Preety good. Great. Terrible. Not bad. 12.Beijing ___ (have)a long history. 13.I am having a great time ____(visit)my grandmother. 14.We feel very ___ after the ___ (relax) vacation. 15.We are ___ vacation in Dalian. A in B on C at Dof 16.What ____ when it’s sunny? Aare you doing B do you do C does you D is he do 17.Tom like to go ___(skate) 18.Kate is very tired, so she needs ____(relax) 19.Would you like to ____ a photofor us? Sure. Agive B make C take Dget 20.Could you help me put up the maps on the wall?_____ A.No problem B. I hope so C. That’s all right D.That’s a good idea. 21.China plan to let Jim ____the Xisha Islands> Visit visits visiting visited 22.Where is Grace? She ____ IN THE yard. Reads read is reading was reading 23.Britain is ___ European country and Singapore is ___ Asian country A,a,a a, an the , an a ,the 战略与策略的主要区别 一，什么是战略营销？ 必须首先明确，什么是战略。 1，战略的本质是一个企业的选择。为什么要做选择？因为任何一个企业都不是全能的。不可能做所有的事情，也不是所有的事情都能做好！任何企业的资源和能力都是有限的。战略就是要把有限的资源和能力，用到产出最大的地方。战略就是一个选择的过程，选择什么？如何选择？这是企业战略规划所要研究的课题。 2，战略首先意味着放弃。在中国目前的经济环境下，战略对于企业家的意义，更为重要的是“放弃”。中国的经济处在快速发展期，有太多的市场机会可供选择。但选择意味着放弃，而放弃是一件很痛苦的事情。 综上所述，战略选择的核心是对企业目标客户群的选择。而战略营销就是从战略的高度思考和规划企业的营销过程，是聚焦最有价值客户群的营销模式。 我们都知道80/20原理，20%的客户创造了企业80%的利润。战略营销要做的就是找到适合企业的目标客户群，并锁定他们进行精确打击，使企业的资源和能力发挥最大的效益，并实现企业能力的持续提升。 因此，战略营销的三个关键要素就是：1）客户细分；2）聚焦客户价值；3）为股东和客户增值。 二，什么是策略营销？ 策略营销主要指的是在市场营销中，将企业的市场策略运用到营销中的过程。 比如： 1，低成本策略 通过降低产品生产和销售成本,在保证产品和服务质量的前提下,使自己的产品价格低于竞争对手的价格,以迅速扩大的销售量提高市场占有率的竞争策略。 2.差别化策略 通过发展企业别具一格的营销活动,争取在产品或服务等方面具有独特性,使消费者产生兴趣而消除价格的可比性,以差异优势产生竞争力的竞争策略。 3.聚焦策略 通过集中企业力量为某一个或几个细分市场提供有效的服务,充分满足一部分消费者的特殊需求,以争取局部竞争优势的竞争策略。 一个企业的市场营销策略必须是在企业的战略营销策略下确定的，可以简单把策略营销理解成企业在市场的战术营销。这就是两者的区别！ with用法归纳 （1）“用……”表示使用工具，手段等。例如： ①We can walk with our legs and feet. 我们用腿脚行走。 ②He writes with a pencil. 他用铅笔写。 （2）“和……在一起”，表示伴随。例如： ①Can you go to a movie with me? 你能和我一起去看电影'>电影吗？ ②He often goes to the library with Jenny. 他常和詹妮一起去图书馆。 （3）“与……”。例如： I’d like to have a talk with you. 我很想和你说句话。 （4）“关于，对于”，表示一种关系或适应范围。例如： What’s wrong with your watch? 你的手表怎么了？ （5）“带有，具有”。例如： ①He’s a tall kid with short hair. 他是个长着一头短发的高个子小孩。 ②They have no money with them. 他们没带钱。 （6）“在……方面”。例如： Kate helps me with my English. 凯特帮我学英语。 （7）“随着，与……同时”。例如： With these words, he left the room. 说完这些话，他离开了房间。 [解题过程] with结构也称为with复合结构。是由with+复合宾语组成。常在句中做状语，表示谓语动作发生的伴随情况、时间、原因、方式等。其构成有下列几种情形： 1.with+名词(或代词)+现在分词 此时，现在分词和前面的名词或代词是逻辑上的主谓关系。 例如:1)With prices going up so fast, we can't afford luxuries. 由于物价上涨很快，我们买不起高档商品。（原因状语） 2)With the crowds cheering, they drove to the palace. 在人群的欢呼声中，他们驱车来到皇宫。（伴随情况） 2.with+名词(或代词)+过去分词 此时，过去分词和前面的名词或代词是逻辑上的动宾关系。 in，on ，at，by ，of ，with 介词区别与用法 in用在年月季节前，还有上午、下午等固定习语里 at用于传统的节日前，如at Christmas等；还有固定词组：at noon, at night;在点时间前用at 如at 7.15 on 用于具体的日期前，星期几，几号，包括那天的上午下午晚上等，如on Friday afternoon with: 一、with表拥有某物 Mary married a man with a lot of money . 马莉嫁给了一个有着很多钱的男人。 I often dream of a big house with a nice garden . 我经常梦想有一个带花园的大房子。 The old man lived with a little dog on the lonely island . 这个老人和一条小狗住在荒岛上。 二、with表用某种工具或手段 I cut the apple with a sharp knife . 我用一把锋利的刀削平果。 Tom drew the picture with a pencil . 汤母用铅笔画画。 三、with表人与人之间的协同关系 make friends with sb talk with sb quarrel with sb struggle with sb fight with sb play with sb work with sb cooperate with sb I have been friends with Tom for ten years since we worked with each other , and I have never quarreled with him . 自从我们一起工作以来，我和汤母已经是十年的朋友了，但我们从没有吵过架。 四、with 表原因或理由 John was in bed with high fever . 约翰因发烧卧床。 He jumped up with joy . 他因高兴跳起来。 Father is often excited with wine . 父亲常因白酒变的兴奋。 五、with 表“带来”，或“带有,具有”，在…身上，在…身边之意 The girl with golden hair looks beautiful . 那个金头发的女孩看起来漂亮。 The famous director will come to the meeting with the leading actor and actress . 介词with的用法大全 With是个介词，基本的意思是“用”，但它也可以协助构成一个极为多采多姿的句型，在句子中起两种作用；副词与形容词。 with在下列结构中起副词作用： 1.“with+宾语+现在分词或短语”，如： (1) This article deals with common social ills, with particular attention being paid to vandalism. 2.“with+宾语+过去分词或短语”，如： (2) With different techniques used, different results can be obtained. (3) The TV mechanic entered the factory with tools carried in both hands. 3.“with+宾语+形容词或短语”，如： (4) With so much water vapour present in the room, some iron-made utensils have become rusty easily. (5) Every night, Helen sleeps with all the windows open. 4.“with+宾语＋介词短语”，如： (6) With the school badge on his shirt, he looks all the more serious. (7) With the security guard near the gate no bad character could do any thing illegal. 5.“with+宾语＋副词虚词”，如： (8) You cannot leave the machine there with electric power on. (9) How can you lock the door with your guests in? 上面五种“with”结构的副词功能，相当普遍，尤其是在科技英语中。 接着谈“with”结构的形容词功能，有下列五种： 一、“with+宾语+现在分词或短语”，如： (10) The body with a constant force acting on it. moves at constant pace. (11) Can you see the huge box with a long handle attaching to it ? 二、“with+宾语+过去分词或短语” (12) Throw away the container with its cover sealed. (13) Atoms with the outer layer filled with electrons do not form compounds. 三、“with+宾语+形容词或短语”，如： (14) Put the documents in the filing container with all the drawers open. 介词“with”的用法 1、同, 与, 和, 跟 talk with a friend 与朋友谈话 learn farming with an old peasant 跟老农学习种田 fight [quarrel, argue] with sb. 跟某人打架 [争吵, 辩论] [说明表示动作的词, 表示伴随]随着, 和...同时 change with the temperature 随着温度而变化 increase with years 逐年增加 be up with the dawn 黎明即起 W-these words he left the room. 他说完这些话便离开了房间。2 2、表示使用的工具, 手段 defend the motherland with one s life 用生命保卫祖国 dig with a pick 用镐挖掘 cut meat with a knife 用刀割肉3 3、说明名词, 表示事物的附属部分或所具有的性质]具有; 带有; 加上; 包括...在内 tea with sugar 加糖的茶水 a country with a long history 历史悠久的国家4 4、表示一致]在...一边, 与...一致; 拥护, 有利于 vote with sb. 投票赞成某人 with的复合结构作独立主格,表示伴随情况时，既可用分词的独立结构，也可用with的复合结构： with +名词(代词)+现在分词/过去分词/形容词/副词/不定式/介词短语。例如： He stood there, his hand raised. = He stood there, with his hand raise.他举手着站在那儿。 典型例题 The murderer was brought in, with his hands ___ behind his back A. being tied B. having tied C. to be tied D. tied 答案D. with +名词(代词)+分词+介词短语结构。当分词表示伴随状况时，其主语常常用 一、with （介词）的复合结构：一做状语，二做定语。 作状语： 1:with +名词/代词，表示随着....， Times change and we must change with them. 时代变了,我们也要跟着变。 The risk of developing heart disease increases with the number of cigarettes smoked. 吸食香烟的数量越多，患心脏病的风险就越大。 Blood pressure decreases with exercise. 血压随着锻炼而降低。 With all his abilities, he failed completely. 尽管很有能力,他还是一败涂地了。 2：with +名词/代词+非谓语动词（现在分词/过去分词/不定式）作伴随状语/原因状语 (1 )with + 名词/代词+V-ing（现在分词） With prices going up so fast, we can't afford luxuries. 由于物价上涨很快，我们买不起高档商品。(原因状语) With the crowds cheering, they drove to the palace. 在人群的欢呼声中，他们驱车来到皇宫。(伴随情况) 注：现在分词和前面的名词或代词是逻辑上的主谓关系，也就是说动作由with 后的名词或代词发出，是主动者 (2)with + 名词/代词+V-ed（过去分词） I sat in my room for a few minutes with my eyes fixed on the ceiling. 我在房间坐了一会儿，眼睛盯着天花板。(伴随情况) She had to walk home with her bike stolen. 自行车被偷，她只好步行回家。(原因状语) 注：过去分词和前面的名词或代词是逻辑上的动宾关系，也就是说with后面的宾语是动作的承受者，是被动的 (3)with或without+名词/代词+ to do (动词不定式) With no one to talk to, John felt miserable. 认清维也纳华尔兹中的重要区别 维也纳华尔兹中的重要区别： 1、左转步与右转步不相同。左转步反身，右转步摆荡； 2、男士步法与女士步法不相同。男士前进摆荡，女士前进无摆荡； 3、前进小节与后退小节不相同。男士前进小节大步向前，后退小节小步调整； 4、节拍长短不相同。每一拍时间值长短不相同，不是平均占一拍。具体来说： 1、维也纳华尔兹左转步与右转步不相同。 在维也纳华尔兹中，右转和左转的跳法是不对称的，右旋转是横并式结构，右转步强调向前流动，强调摆荡，有倾斜，有起伏，步幅大，以单侧拉腰为主；左旋转是锁式结构，左转步强调拧腰胯，反身，无摆荡，无升降，锁步，步幅小，要不停地反身。 2、维也纳华尔兹男士步法与女士步法不相同。 在维也纳华尔兹中，男士与女士步法不相同，男士的前进转身小节是女士的后退转身小节，男士前进右转摆荡，女士后退右转也摆荡；男士后退右转无摆荡，女士前进右转也无摆荡。 3、维也纳华尔兹前进小节与后退小节跳法不相同。 在维也纳华尔兹中，前进与后退小节跳法不相同，男士前进小节大步向前，后退小节小步调整。右转男士后退（女士前进）那个小节不摆荡，步子也较小，相当于休息。 4、维也纳华尔兹中节拍长短不相同。 在维也纳华尔兹中，每两小节六步为一组，每一节拍时间值长短不相同，不都是平均占一拍。六个节拍时间值分别是：1.5、0.75、0.75、1.5、0.75、0.75，第一、四拍最长，第三、六拍最短，口令：慢、快、快、慢、快、快 跳快三的要领 (2011-11-14 10:18:09) 转载▼ 标签： 杂谈 维也纳华尔兹俗称快三，它是舞中之王，跳快三是很难跳得好的，我虽然跳舞多年，长期以来被错误的观点支配，也是最近才掌握到跳快三的要领。 快三看似简单，只有四种基本步法，左转、右转、左换步、右换步，但如果不掌握要领，光靠看视频，听舞友指点，不容易领会关键的要领，舞就跳不好。 很多人以为快三就是比慢三转快一些，这就错了，这也是跳不好快三的原因。我以前也是用这种思维去跳的，结果一直转不好，转起来不畅顺，不能绕舞池转。开始还以为对方没跳好，但与多个人跳过也不好，最近才发现，是自己没跳对，不会带舞伴，跳和带的方法不对。 不久前，在网上无意间找到了2句跳好快三的要决，原来快三的转与慢三的转完全不一样，慢三是转园圈，像车轮那样。而快三的转是折转或翻转，像蛇爬行时一样。其次，快三不是以3拍为一小节，而是以6拍为一小节。一小节中跳半个大圈和半个小圈，不是两个半圈相同的，跳时男女互相错开，男跳大半圈时女跳小半圈，只有跳大半圈时才发力。这就是对快三的新认识，是跳快三的要领。 从以上认识入手，还需要学会用力的方法，以前我和很多人一样用手发力来带对方转，这显然不能到位。其实，关键是要从腰发力，用侧腰的力去带动身体前进，以前进带动转动。 快三的左转和右转也很不一样，很多人右转不错，但左转就不妥，这也是上面说的原因，没有认识快三的转的实质。在跳快三过 英语介词with的用法 with结构是许多英语复合结构中最常用的一种。学好它对学好复合宾语结构、不定式复合结构、动名词复合结构和独立主格结构均能起很重要的作用。本文就此的构成、特点及用法等作一较全面阐述，以帮助同学们掌握这一重要的语法知识。 一、 with结构的构成 它是由介词with或without+复合结构构成，复合结构作介词with或without的复合宾语，复合宾语中第一部分宾语由名词或代词充当，第二部分补足语由形容词、副词、介词短语、动词不定式或分词充当，分词可以是现在分词，也可以是过去分词。With结构构成方式如下： 1. with或without-名词/代词+形容词； 2. with或without-名词/代词+副词； 3. with或without-名词/代词+介词短语； 4. with或without-名词/代词+动词不定式； 5. with或without-名词/代词+分词。 下面分别举例： 1、 She came into the room，with her nose red because of cold.（with+名词+形容词，作伴随状语） 2、 With the meal over ， we all went home.（with+名词+副词，作时间状语） 3、The master was walking up and down with the ruler under his arm。（with+名词+介词短语，作伴随状语。） The teacher entered the classroom with a book in his hand. 4、He lay in the dark empty house，with not a man ，woman or child to say he was kind to me.（with+名词+不定式，作伴随状语） He could not finish it without me to help him.（without+代词 +不定式，作条件状语） 5、She fell asleep with the light burning.（with+名词+现在分词，作伴随状语） 6、Without anything left in the cupboard， she went out to get something to eat.（without+代词+过去分词，作为原因状语）英语介词用法大全
介词in,on.at,for.with,by,of的基本用法
by with and in区别
compare 的两个重要词组区别
with的用法大全
中考介词in和with 用法
战略与策略的主要区别
with用法归纳
in,on ,at, by ,of ,with 介词区别与用法
介词with的用法大全
初中 英语 介词“with”的用法
as和with区别
认清维也纳华尔兹中的重要区别
英语介词with的用法