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Branch and Bound Methods

Branch and Bound Methods
Branch and Bound Methods

Branch and Bound Methods

Stephen Boyd and Jacob Mattingley

Notes for EE364b,Stanford University,Winter2006-07

March11,2007

Branch and bound algorithms are methods for global optimization in nonconvex prob-lems[LW66,Moo91].They are nonheuristic,in the sense that they maintain a provable upper and lower bound on the(globally)optimal objective value;they terminate with a certi?cate proving that the suboptimal point found is?-suboptimal.Branch and bound al-gorithms can be(and often are)slow,however.In the worst case they require e?ort that grows exponentially with problem size,but in some cases we are lucky,and the methods converge with much less e?ort.In these notes we describe two typical and simple examples of branch and bound methods,and show some typical results,for a minimum cardinality problem.

1Unconstrained nonconvex minimization

The material in this section is taken from[BBB91].The branch and bound algorithm we describe here?nds the global minimum of a function f:R m→R over an m-dimensional rectangle Q init,to within some prescribed accuracy?.We let f?denote the optimal value, i.e.,f?=inf x∈Q

f(x).For a rectangle Q?Q init we de?ne

init

f(x),

Φmin(Q)=inf

x∈Q

so f?=Φmin(Q init).

The algorithm will use two functionsΦlb(Q)andΦub(Q)de?ned for any rectangle Q?Q init.These functions must satisfy the following conditions.First,they are lower and upper bounds onΦmin(Q),respectively:for any Q?Q init,

Φlb(Q)≤Φmin(Q)≤Φub(Q).

The second condition is(roughly)that the bounds become tight as the rectangle shrinks to a point.More precisely,as the maximum half-length of the sides of Q,denoted by size(Q), goes to zero,the di?erence between upper and lower bounds uniformly converges to zero, i.e.,

??>0?δ>0?Q?Q init,size(Q)≤δ=?Φub(Q)?Φlb(Q)≤?.

1

Figure1:Branch and bound example in R2,after3iterations.The partition of

the original rectangle is shown at left;the associated binary tree is shown at right.

We mention a third condition,which is not needed to prove convergence of the branch and bound algorithm,but is needed in practice:The functionsΦub(Q)andΦlb(Q)should be cheap to compute.

We now describe the algorithm.We start by computing

L1=Φlb(Q init),U1=Φub(Q init),

which are lower and upper bounds on f?,respectively.If U1?L1≤?,the algorithm terminates.Otherwise we partition Q init into two rectangles Q init=Q1∪Q2,and compute Φlb(Q i)andΦub(Q i),i=1,2.Then we have new lower and upper bounds on f?,

L2=min{Φlb(Q1),Φlb(Q2)}≤Φmin(Q init)≤U2=min{Φub(Q1),Φub(Q2)}.

If U2?L2≤?,the algorithm terminates.Otherwise,we partition one of Q1and Q2into two rectangle,to obtain a new partition of Q init into three rectangles,and computeΦlb and Φub for these new rectangles.We update the lower bound L3as the minimum over the lower bounds over the partition of Q init,and similarly for the upper bound U3.

At each iteration we split one rectangle into two,so after k iterations(counting the evaluation ofΦlb(Q init)andΦub(Q init)as the?rst one),we have a partition of Q init into k rectangles,Q init=∪k i=1Q i.This partition can be represented by a partially?lled binary tree,with the root node corresponding to the original rectangle,and the children of a node corresponding to the subrectangles obtained by partitioning the parent rectangle.

This is illustrated in?gure1.The original rectangle corresponds to the root of the tree.We?rst partition Q init vertically.At the next iteration,we horizontally partition the lefthand child;at the third iteration,we vertically partition the bottom child.After three iterations,we have a partition of the original rectangle into4rectangles,which correspond to the leaves in the tree.

The associated lower and upper bounds on f?are

L k=min

i=1,...,k Φlb(Q i),U k=min

i=1,...,k

Φub(Q i).

2

We can assume that the lower and upper bounds of the pair of rectangles obtained by splitting are no worse than the lower and upper bounds of the rectangle they were formed from.This implies that L k is nondecreasing,and U k is nonincreasing.

To fully specify the algorithm,we need to give the rule for choosing which rectangle to split at each step,and we have to specify which edge along which the rectangle is to be split. One standard method for choosing the rectangle in the current partition to be split is to choose one with the smallest lower bound,i.e.,a rectangle that satis?esΦlb(Q)=L k.Once we choose the rectangle to split,we split it along any of its longest edges.

The algorithm is summarized below.

k=0;

L0={Q init};

L0=Φlb(Q init);

U0=Φub(Q init);

while U k?L k>?,{

pick Q∈L k for whichΦlb(Q)=L k;

split Q along one of its longest edges into Q I and Q II;

form L k+1from L k by removing Q k and adding Q I and Q II;

L k+1:=min Q∈L

Φlb(Q);

k+1

Φub(Q);

U k+1:=min Q∈L

k+1

k:=k+1;

}

The requirement that we split the chosen rectangle along a longest edge controls the condition number of the rectangles in the partition,which in turn allows us to prove conver-gence.

As the algorithm proceeds,we can eliminate some rectangles from consideration;they can be pruned,sinceΦmin(Q init)cannot be achieved in them.This is done as follows.At each iteration,we eliminate from the list L k any rectangles that satisfyΦlb(Q)>U k,since every point in such a rectangle is worse than the current upper bound on f?.

Pruning does not a?ect the algorithm,since any rectangle pruned would never be selected later for further splitting.But pruning can reduce storage requirements.With pruning,the union of the list of current rectangles can be regarded as the set of possible minimizers of f.The number of active rectangles,and their total volume,give some measure of algorithm progress.

The term pruning comes from the following.The algorithm can be viewed as developing a partially?lled binary tree of rectangles representing the current partition L k,with the nodes corresponding to rectangles and the children of a given node representing the two halves obtained by splitting it.The leaves of the tree give the current partition.By removing a rectangle from consideration,we prune the tree.

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1.1Convergence analysis

We?rst show that after a large enough number of iterations,the partition L k must contain a rectangle of small volume.We then show that this rectangle has small size,and this in turn implies that U k?L k is small.

The number of rectangles in the partition L k is k,and the total volume of these rectangles is vol(Q init),so

min Q∈L k vol(Q)≤

vol(Q init)

min i(u i?l i)

.

Our splitting rule,which requires that we split rectangles along a longest edge,results in an upper bound on the condition number of rectangles in our partition:for any rectangle Q∈L k,

cond(Q)≤max{cond(Q init),2}.(2) To prove this it is enough to show that when a rectangle Q is split into rectangles Q1and Q2,we have

cond(Q1)≤max{cond(Q),2},cond(Q2)≤max{cond(Q),2}.

Letνmax be the maximum edge length of Q,and letνmin be the minimum edge length of Q, so cond(Q)=νmax/νmin.When Q is split into Q1and Q2,the maximum edge length of the children is no more thanνmax.The minimum edge length of the children,by our splitting rule,is no smaller than the minimum ofνmax/2andνmin.It follows that the condition number of the children is no more than the maximum of2and cond(Q).

We note that there are other splitting rules that also result in a uniform bound on the condition number of the rectangles in any partition generated.One such rule is to cycle through the index on which we split the rectangle.If Q was formed by splitting its parent along the i th coordinate,then when we split Q,we split it along the(i+1)modulo m coordinate.

We can bound the size of a rectangle Q in terms of its volume and condition number, since

vol(Q)= i(u i?l i)

≥max

i

(u i?l i) min i(u i?l i) m?1

=

(2size(Q))m

cond(Q) m.

4

Thus,

1

size(Q)≤

max{cond(Q init),2} vol(Q init)

2

We can also get an upper bound on p?using the relaxation.The simplest method is to round each of the relaxed Boolean variables z i to0or1.A more sophisticated approach is to?rst round each of the relaxed Boolean variables z i to0or1,and then,with these values of z i?xed,solve the resulting convex problem in the variable x.

The upper bound obtained from either of these methods can be+∞,if this method fails to produce a feasible point.(The upper bound can also be?∞,in which case the original problem is surely unbounded below.)We’ll denote this upper bound by U1.Of course,if we have U1?L1≤?,the required tolerance,we can quit.

Now we are going to branch.Pick any index k,and form two problems:The?rst problem is

minimize f0(x,z)

subject to f i(x,z)≤0,i=1,...,m

z j∈{0,1},j=1,...,n

z k=0

and the second problem is

minimize f0(x,z)

subject to f i(x,z)≤0,i=1,...,m

z j∈{0,1},j=1,...,n

z k=1.

In other words,we?x the value of z k to0in the?rst problem,and1in the second.We call these subproblems of the original,since they can be thought of as the same problem, with one variable eliminated or?xed.Each of these subproblems is also a mixed Boolean convex problem,with n?1Boolean variables.The optimal value of the original problem is the smaller of the optimal values of the two subproblems.

We now solve the two convex relaxations of these subproblems to obtain a lower and upper bound on the optimal value of each subproblem.We’ll denote these as?L,?U(for z k=0) andˉL,ˉU(for z k=1).Each of these two lower bounds must be at least as large as L1,i.e., min{?L,ˉL}≥L1.We can also assume,without loss of generality,that min{?U,ˉU}≤U1. From these two sets of bounds,we obtain the following bounds on p?:

L2=min{?L,ˉL}≤p?≤U2=min{?U,ˉU}.

By the inequalities above,we have U2?L2≤U1?L1.

At the next step,we can choose to split either of the subproblems.We need to split on an index(not equal to k,the index used to split the original problem).We solve the relaxations for the split subproblems(which have n?2Boolean variables),and obtain lower and upper bounds for each.

At this point we have formed a partial binary tree of subproblems.The root is the original problem;the?rst split yields two child subproblems,one with z k=0and the other with z k=1.The second iteration yields another two children of one of the original children. We continue is this way.At each iteration,we choose a leaf node(which corresponds to a subproblem,with some of the Boolean variables?xed to particular values),and split it,by

6

?xing a variable that is not?xed in the parent problem.An edge in the tree corresponds to the value0or1of a particular variable z i.At the root node of the tree,none of the values of the z i are speci?ed.A node at depth k in the tree corresponds to a subproblem in which k of the Boolean variables have?xed values.For each node,we have an upper and lower bound on the optimal value of the subproblem.After k iterations,we have a tree with exactly k leaves.

The minimum of the lower bounds,over all the leaf nodes,gives a lower bound on the optimal value p?;similarly,the minimum of the upper bounds,over all the leaf nodes,gives an upper bound on the optimal value p?.We refer to these lower and upper bounds as L k and U k,respectively.The global lower bound L k is nondecreasing,and the global upper bound U k is nonincreasing,so we always have U k+1?L k+1≤U k?L k.We can terminate the algorithm when U k?L k≤?.

Proving convergence of the algorithm is trivial:it must terminate,with U k=L k,in fewer than2n iterations.To see this,note that if a leaf has depth n,it means that all the Boolean variables are?xed in the subproblem,so by solving the convex relaxation we get the exact solution.In other words,for any leaf of depth n,we have U=L.The worst thing that can happen is that we develop a complete binary tree,to depth n,which requires2n iterations,at which point every subproblem lower and upper bound is equal,and therefore the algorithm terminates.This is nothing more than exhaustive search.

We can carry out pruning,to save memory(but not time),as the algorithm progresses.At iteration k we have the upper bound U k on p?.Any node in the tree that has a lower bound larger than U can be pruned,i.e.,removed from the tree,since all points corresponding to that node are worse than the current best point found.

The choice of which leaf to split at a given stage in the algorithm is somewhat arbitrary. The primary strategy is to split the leaf corresponding to the smallest lower bound.This means we always get an improvement in the global lower bound.The drawback of this scheme is that in some situations(such as the cardinality example in the next section)we may achieve a tight lower bound before getting any optimal feasible points.

The same can be said for the choice of index to use to split a leaf.It can be any index corresponding to a Boolean variable that has not yet been?xed.One heuristic is to split along a variable k for which|z?k?1/2|is smallest,where z?k is the optimal point found in the relaxation.This corresponds to splitting a variable that is most‘undecided’(between 0and1)in the relaxation.In some problems,the opposite heuristic works well:we split a variable that is most‘decided’.To do this,we choose k for which|z?k?1/2|is maximum. In most cases,many of the variables z?k will be0or1;among these,we can choose the one with the largest associated Lagrange multiplier.(The idea here is that this is the relaxed variable that is0and1,and has the highest pressure to stay there.)

7

After this third iteration we have

L3=min{∞,∞,1}=1,U3=min{∞,∞,1}=1,

which means we have solved the problem globally.

Of course for a simple problem like this we could have simply checked the8possible Boolean vectors for feasibility,and evaluated the objective function of those that are feasible. This example was meant only to illustrate the branch and bound method.

3Minimum cardinality example

We consider the problem of?nding the sparsest solution of a set of linear inequalities,

minimize card(x)

(7)

subject to Ax b.

This problem is equivalent to the mixed Boolean-LP

minimize1T z

subject to L i z i≤x i≤U i z i,i=1,...,n

(8)

Ax b

z i∈{0,1},i=1,...,n

with variables x and z,and where L and U are,respectively,lower and upper bounds on x. We can?nd the lower bounds L i by solving the n LPs

minimize x i

subject to Ax b,

for i=1,...,n.(The optimal values of these LPs give L1,...,L n.)We can?nd the upper bounds U1,...,U n by solving the n LPs

maximize x i

subject to Ax b

for i=1,...,n.

If any L i is positive,then any feasible x satis?es x i>0,and it follows that z i=1.In this case,we can remove the variable z i(since we already know its value).In a similar way,if any U i is negative,then we must have z i=1,and we can simple remove z i.So we’ll assume that L i≤0and U i≤0.

We’ll solve the mixed Boolean-convex problem(8)using the general method described in§2.The relaxation is

minimize1T z

subject to L i z i≤x i≤U i z i,i=1,...,n

(9)

Ax b

0≤z i≤1,i=1,...,n

9

which is equivalent to

minimize n i=1((1/U i)(x i)++(?1/L i)(x i)?)

subject to Ax b

(assuming L i<0and U i>0,for i=1,...,n).The objective here is an asymmetric weighted ?1-norm.Minimizing the?1-norm over a convex set is a well known heuristic for?nding a sparse solution.We can improve on the lower bound found by the relaxation.Since card(x) is integer valued,if we know card(x)≥l,we know that card(x)≥?l?,the smallest integer greater than or equal to l.(For example,if we know that card(x)≥23.4,then in fact we know that card(x)≥24.)

We?nd an upper bound on the mixed Boolean-LP(8)by simply recording the cardinality of the x obtained in the relaxed problem(9).

At each iteration we split a node with the lowest lower bound.This will be the optimal strategy if we?nd an optimal point before proving the lower bound.Less clear is the choice of variable to split;empirical results suggest that splitting along the‘most ambiguous’variable in the relaxation,i.e.,z j,where j=argmin i(|z i?1/2|),is an e?cient strategy.

3.1Numerical results

A small example

We?rst give the results of the branch and bound algorithm for a randomly generated problem of the form(7),with x∈R30and100constraints.It takes124iterations,and the solution of a total of309LPs(including the60LPs required to compute the lower and upper bounds on each variable)to?nd that the minimum cardinality is19.

Figure3shows the tree generated by the algorithm at four di?erent iterations.At top left is the tree after three iterations,with lower bound12.6and upper bound21.At top right is the tree after?ve iterations,with lower bound13.1and upper bound20.Note that there is a pruned node,shown with a dotted line.It had a lower bound above20,the global upper bound at that point,and so could be eliminated from the tree.At bottom left,after 10iterations,the lower bound is13.9and upper bound19.Finally,the plot at bottom right shows the tree at the end of the algorithm,with lower bound18.04and upper bound19.

Figure4shows the evolution of the global lower and upper bounds.We can see,for example,that a point with globally minimum cardinality is obtained after only8iterations; however,it takes around100more iterations to guarantee that this point has minimum cardinality.

Figure5shows the portion of non-eliminated or non-pruned sparsity patterns,versus iteration number.Each pruned node,at level i,corresponds to2n+1?i possible values of the Boolean variable z;each active(non-pruned)node at level i corresponds to2n+1?i possible values of the Boolean variable.Near termination,the fraction of non-pruned Boolean values is around1%.This means we have eliminated around99%of the possible Boolean values. However,the total number of Boolean values is230≈109,so1%still represents around107 possible values for z.This means that,right before the algorithm terminates,there are still

10

Figure3:Tree with30variable problem,after3iterations(top left),5iterations (top right),10iterations(bottom left),and124iterations(bottom right).

11

107possible values of z‘in play’.

Figure6shows the total number of active(non-pruned)leaves in the tree,which is the size of the partition,versus iteration number.After k iterations,we can have at most k leaves(which occurs only if no pruming occurs).So we can see that in this example,we have typically pruned around half of the leaves.

For this example,direct enumeration of all sparsity patterns is impractical(or at least, very slow compared to branch and bound),since it would require the solution of230≈109 LPs.We must count ourselves lucky to have solved the problem with(provable)global optimality by solving only a few hundred LPs.

14

Acknowledgments

Arpita Ghosh and Alessandro Magnani helped with an earlier version of this set of notes. References

[BBB91]V.Balakrishnan,S.Boyd,and S.Balemi.Branch and bound algorithm for com-puting the minimum stability degree of parameter-dependent linear systems.Int.

J.of Robust and Nonlinear Control,1(4):295–317,October–December1991. [LW66] https://www.sodocs.net/doc/1d8214428.html,wler and D.E.Wood.Branch-and-bound methods:A survey.Operations Research,14:699–719,1966.

[Moo91]R.E.Moore.Global optimization to prescribed https://www.sodocs.net/doc/1d8214428.html,puters and Mathe-matics with Applications,21(6/7):25–39,1991.

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公元前204年,汉王刘邦派大将韩信和张耳,率领汉军去攻打赵国。赵王和赵军统帅陈余,率领二十万兵马,集结在井陉(xíng),准 备迎战。 赵国谋士李左车向陈余献计说:“我有个主意,井陉口这个地方 两旁有山,道路狭窄,车马很难通过。我们派三万人抄后路截断他们 的粮车,你统率大军正面阻击汉军,把城墙垒得高高的,不与汉军交战。这样一来,汉军前进不得,后退不能,又无粮草,成了瓮(wèng) 中之鳖(biē)。我看不出十天,我们就能够捉住韩信。”但是,赵王没 有采纳李左车的意见。 韩信知道李左车的计策没有被采用,十分高兴。他把兵马集结在 离井陉口三十余里的地方。到了后半夜,韩信派两千名轻骑兵,每人 带一面汉军红旗,从小路迂回到赵营的侧后方,埋伏起来。命令他们 待赵军倾巢而出时,袭入其大营,拔去赵军旗帜,全部插上汉军的红旗。韩信又派一万人马作先头部队,沿着河岸摆开阵势。 陈余等人看见韩信把兵马安置于背水之处,大笑他是傻瓜,不懂 用兵,便率领赵兵全力迎战。汉军的二千轻兵见赵军全营出动,只留 下一个空营,立即闯进赵营,拔掉赵旗,换上汉旗。韩信则假装败退,向河岸阵地退去。 赵军追到汉军靠河阵地,汉军后无退路,个个背水拼命厮杀。而 赵军久战不胜,士气开始低落,又发现背后自己的军营都插上了汉军 的旗帜,军心顿时大乱,纷纷溃(kuì)逃。汉军乘机前后夹攻,大破赵军。 “背水一战”这则成语的意思是比喻决一死战。 历史典故的四字成语(四):四面楚歌 项羽和刘邦原来约定以鸿沟(在今河南荣县境贾鲁河)东西边作为 界限,互不侵犯。后来刘邦听从张良和陈平的规劝,觉得应该趁项羽 衰弱的时候消灭他,就又和韩信、彭越、刘贾会合兵力追击正在向东

allegro使用技巧

为了便于大家察看pcb 版,我将Allegro 中遇到的一些细微的东西在此跟大家分享: 1、焊盘空心、实心的显示 经常每个人都有自己视觉的习惯,有些人习惯空心焊盘而有些人则习惯实心的,当面对的板子和你自己的习惯矛盾时,可以用以下的方法来改变: 在菜单中选SetupÆDrawing Options….,会弹出一个对话框:在Display 下的Filled pad 前面打勾,显示的就是实心焊盘,反之就是空心的。 在16.3中则在display菜单下参数设置,display选项卡中 2、Highlight 这个如果没有设定好的话,当我们高亮一个网络或者零件的时候,显示为虚线条,这样当放大屏幕的时候很难看清点亮的东西。没有设定好的话,当我们高亮一个网络或者零件的时候,显示为虚线条,这样当缩小屏幕的时候很难看清点亮的东西。按照如下的方法可以加以设定: 在菜单中选SetupÆUser Preferences…,点选Display,在右侧的Display_nohilitefont 前面打勾,则高亮的物体显示为实心颜色,否则为虚线。这一点实际做一下对比就可以体会到。 3、显示平面层花盘 这点跟第1 点类似,在图一中的Thermal pads 中打勾即可;另外要想显示钻孔,只需选中Display drill holes。 4、DRC 显示为填充以及改变大小 显示填充:同样在图二的对话框中,选中右侧Display_drcfill 即显示填充的drc,否则为空心。 改变大小:在参数设置中显示的对话框中点开drc 则出现对话框: 我们就可以更改drc 的大小,或者开、关drc。 5、改变光标的形状(大十字、小十字等) 用惯PowerPCB 的人可能比较习惯光标是大十字,充满整个屏幕,可以作如下设定: 在图二中,选中左侧Ui,在右侧Pcb_cursor 的下拉菜单中选不同的项,则可以实现不同的设定,其中Cross 是小光标,infinite 是大光标。 6、将整版显示为0mil 的线宽 选中右侧nolinewith 可以实现。 7、动态的显示布线长度 在图二的对话框中选中左侧的Etch,右侧选中Allegro_etch_length_on,这样在布线的时候就可以实时的显示已布线的长度,当然并不是所有时候都方便,有时候可能后觉得碍眼,看情况了。 以上是我已发现的一些东东,不对指出还往指正。这些都是很细节的问题,知道了可能会觉得很简单,不知道的话怎么找也找不着,当然还有很多没有发现的东西,如果你已经发现了,麻烦你告诉我一声,我再有什么发现的话还会继续与大家分享。 1.ALLEGRO 自动布线后,为直角调整成45度角走线: Route-Gloss-Parameters-Convert corner to arc。

英语中常见的同义词(组)集锦

表示“需要”(v)的单词:need, require, involve ,entail 表示“意味着”的单词:mean, involve, entail 表示“宿舍”的单词:dormitory, quarters, lodgings 表示“住宿”的单词:accommodations, (a)lodging 表示“欺骗”(v)的单词:cheat, deceive, trick, fool 表示“渴望”(v)的单词和词组:desire, yearn for/after, long for, thirst for,aspire to do sth, look forward to doing sth, keen on/about doing sth, keen to do sth, keen that, have an urge to do sth (具有做某事的强烈欲望,非常想做某事,形容极度渴望) 表示“同情”(v)的词组:sympathize with, yearn for, feel sympathy for 表示“重要的”的单词:important, crucial, vital, essential, integral, indispensable, significant 表示“参加,从事”的词组:take part in, join(in),participate in, involve sb in(主动, 其被动形式为sb be involved in), sb be engaged in=sb be employed in 表示“能力”的单词:ability和capability(主要用于人)talent(只用于人), competence, capacity(可用于人或物), faculty 表示“想象”(v)的单词和词组:imagine, fancy, fantasy,envisage(英)=envision(美) conceive of 表示“包含(v)”的单词和词组:include, involve, comprehend, inclusive of, embrace, embody, comprise, contain(实际包含的量),encompass 表示“能容纳”的词组:can hold/accommodate 表示“表现,体现”(v)的单词:render, reflect, display, indicate, manifest, embody 表示“导致,引起(v)”的单词和词组:render, entail, lead to, result in, bring about, cause, trigger=prompt(诱发),pose(本义为造成问题或危机) 表示“强大的,强有力的”的单词和词组:powerful, mighty, potent, strong 表示“(感觉等)敏锐的,灵敏的”的单词:keen, acute(首选这两个词)perceptive 表示“提供”的单词和词组:provide/supply sb with sth=provide/supply sth for sb=render/offer sb sth, contribute sth to sth 表示“忙于……,埋头于……”的单词和词组:be buried in, be employed in, be engaged in, sb be immersed in=immerse oneself in 表示“致力于”的词组:devote oneself to =dedicate oneself to, sb be devoted to=sb be dedicated to 表示“表现,体现”(v)的单词:render, reflect, manifest, characterize 表示“显示,表明”(v)的单词:show, suggest, indicate, reflect, reveal,(首选这三个词)emerge 表示“证明”(v)的单词:validate, demonstrate, verify, confirm,prove 表示“证实(v)”的单词:affirm,confirm, prove 表示“说明,阐释(v)”的单词:explain, illustrate, elaborate 表示“持续不断的”的单词:continual,continuous, incessant 表示“(计划,规定等)被实施,生效”的词组有come into effect/force, be implemented, be put into effect/execution, be carried into execution=be put in/into execution 表示“意识到+从句”的词组:know+that从句realize+that从句be/become aware of the fact+ that从句be conscious+that 从句 表示“阻止,阻碍”(v)的单词或词组:prevent/inhibit/hinder sb from doing sth, impede sb to do sth, hamper sb in doing sth 表示“强烈的,剧烈的”的单词:severe, intense, violent, poignant, acute,keen 表示“严重的”的单词:serious, severe, acute, grave 表示“著名的,闻名的”的单词:famous(一般用于褒义),eminent,

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