Lecture 10

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Lecture 10汉译英谓语的处理

Lecture 10
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汉译英中谓语的处理
Lecture 9
After-class exercises
• 8、这时寂静无声。 • silent / silence • 参考译文：It is very quiet at the moment. • 9、屋里有一位客人。 • 参考译文：There is a guest in the room.
• 2、这儿将修建更多的居民大楼。 • There will build more apartment buildings. • Many more apartment buildings will be constructed here. • 参考译文：More apartment buildings will be built here.
Lecture 10
汉译英中谓语的处理
• ―You mean me? Well, yes, I think I could make people dance to my tune. But today it’s another story. It’s not me but others who have changed. People’s moods change everyday.‖
Lecture 9
After-class exercises
• 11、只有慎之又慎，才能避免犯不必要的错误。 • The more cautions you pay, the less mistakes you will make. • Only if we being very careful, can we get rid of some errors. • 参考译文：Being very careful, you (we) can avoid making unnecessary mistakes.

ACM(lecture_10)特殊的数

空间换时间！！
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Catalan number
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HDOJ_1134: Game of Connections
8 7 6
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1 2 3
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Catalan Number
Catalan numbers (1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, ...)
trees with n+1 nodes
3 nodes:
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4 nodes: 5 nodes:
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4. the number of paths of length 2n through an n-
by-n grid that do not rise above the main diagonal
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January
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February
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March
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April
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May、June…
???
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The number series is—— 1、1、2、3、5… This is fibonacci number！
Some other pictures
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Lecture 10

mvcos
m:光学系统的倍率
θ:为点光源的像的移动方向与感光元件方向之间的夹角
感光元件方向上的速度 v cos
可以按 v cos pf 。f 为频率。
m
应用：
(1)在汽车速度检测方面使用该传感器可得到更精确的结果。
(2) 在泥石流和河流等流速的测定方面。
4.3.3 利用相关法的速度传感器
m：质量块的质量 k：弹簧的弹性常数 c：阻尼器的粘性阻尼系数
x m
检测仪 xr=x-x1
k
c
x1
在ζ=0.7附近,λ<<1(振动频率远远小于固有频率)，增益|G(jω)| 为这检倍时测，1n2检物所，测体以相到振可位的动以差相加用近对速于似位度测为移幅试值物1xr8的体a0°1幅的的。值振1n即2是动加速度。
惠斯登电桥如图1所示。被测电阻与三个已知电阻R1、 R2、R0连成电桥的四个臂。四边形的对角线AC接电源，称为电桥的电源对角线；另一对角线BD接检流计G，称为电桥的检测对角线，也称为“桥路”。G的作用就是检测B、D这两点间的电位差。
R1 I1 A
I2 R2
I
B
G Ig
Kg
D
Rx Ix C I0
钢板的移动速度v可由式 v L 得到。
0
4.3.4 多普勒速度计
多普勒效应可用于速度计，从信号发射源向运动速度为v的物体发射
频率为f0的电波或者声波，若在信号发射源(观测点)接受到从运动物体反
射回来的信号频率为fr，则多普勒频率 fd=fr- f0=2 f0
运动物体的速度为
v

2
cf d
f 0 cos
4.4.2 压电元件 (力传感器)

lecture10

第10讲：递归计算机与工程问题解决导论Prof.Steven R.LermanDr.V.Judson Haward方法和变量，1方法和变量，栈不同的变量i会相互影响吗？递归：一个方法可以调用自身吗？栈上的递归方法递归•递归算法获取一个问题并试图以更小的版本或与问题类型相同的版本表达－factorial( n ) = n * factorial( n-1 );•递归算法必须有终止条件或基本情况，这样它就能直接计算答案－factorial( 1 ) = 1•电脑然后解开先前的对栈的调用使用递归算法合并结果－factorial( n ) = n * factorial( n-1 );•3个步骤：检查（1）分割问题，（2）识别基本情况/终止条件，（3）合并结果递归，2•使用递归的原因不是它特别有效，而是常常使用递归比较清晰－可以像使用迭代那样轻松的计算阶乘递归，第二个例子•我们能迭代地得到一个整数数组的最大值吗？•我们可以用递归得到吗？－我们如何分割问题？－基本情况/终止条件是什么？－我们如何来合并结果？递归，第二个例子找到一个整数数组的最大值，1递归，第二个例子找到一个整数数组的最大值，2找到一个整数数组的最大值，3•注意到有两个版本的maxArray()，接受所有的数组作为参数的初步的版本和递归版本•这个方法如何分割问题？－第13行是关键的。

确信它正确•哪一个执行基本情况/终止条件？•哪一行合并结果?取幂•简单取幂是无效率的。

－x的y次方即为y-1个x相乘•例如，x7 = x * x * x * x * x * x * x－连续的平方更有效率，但是实施的时候需要更加注意－例如，x48 = ((((x * x * x)2) 2) 2) 2 用6次乘法运算代替48－使用连续开方得到x的y次方需要多少次乘法运算？•提示：为了得到x1,000,000,000 ，平方只需要30次运算，而简单方法需要1，000，000，000次!取幂（续）•奇数则更加复杂些：－x7 = x * (x*x*x) 2使用4次运算而不是7次－x9 = x * (x*x)2 )2 使用4次运算而不是9次•我们可以概括这些资料，设计出一种算法能够使用二次幂来快速取幂•编写叠代，记录奇数和偶数将会很有用•这里很自然地便用到取幂算法－编写3个恒等方程的系列并把它们作为Java函数完善补充取幂（续）•3个恒等方程：－x1= x（足够小）－x2n= xn* xn（分解问题）－x2n+1= x* x2n(分解问题)递归取幂递归取幂，2•我们为什么使用BigInteger代替标准long算法？•基本情况/终止条件是什么？•有多少种方法来分割问题？•结果是如何合并的？递归取幂，main()递归和迭代•这是一个灵活的关于取幂循环的练习－若有时间和兴趣可以尝试•通常更容易得到一个正确的递归执行－递归通常更接近于基础数学•编译人员和算法设计人员使用一种机械的方法将递归转化为迭代。

lecture 10

价格看涨时,获得期权费利益(部分)
价格看跌时,从IPO机构投资者手中按行使价购买股票
雷曼兄弟招商银行股票挂钩结构存款

投资最长周期: 6个月 (最短1个月) 挂钩股票: 中国招商银行(3968.HK) 最低投资额:HK$500,000/US$65,000 每交易单面值(denomination): HK50,000/US$5,000/每单潜在回报率(Coupon):
股权房地产资产证券化：REIT
房地产资产作为标底资产的证券化种类

股权信托

拥有和管理能够产生现金流的房地产

投资或出售房地产租赁房地产开发物业管理

抵押信托

直接向房地产的所有者或开发商提供信贷,房地产资产组合所获得的租金和贷款利息收入为投资者收益间接购买房地产抵押证券股权和抵押REIT结合
雷曼兄弟招商银行股票结构存款

参考股价(reference price): 敲定日估价时间价格行使价: 94%参考股价优惠券设定的股价(票息价): 105%参考股价赎回时股价: 105%参考股价提前赎回: 如股票在各观察期的最后一天接近或超过相应的赎回价, 存单将会按100%本金和相应的利率收入赎回. 到期赎回: (1) 如果到期股价接近或超过行使价, 投资者收到100% 本金和相应利息 (2)如果到期股价接近或低于行使价, 投资者收到按行使价计算的股票数外加相应利率收入
第十讲：企业上市与资产证券化
曹勇新加坡，南洋理工大学
公司股权的私募与公开发售的区别

私募将公司股份卖给少数策略投资者，如，风投与私募基金，富裕人士(angels天使)

lecture10

NP 完全性的证明SAT限制法局部替换法构件设计法恰好覆盖3SAT 最大可满足性构件设计法构件设计法构件设计法子集和VC 有向HC 限制法局部替换法HC独立集双机调度0-1背包局部替换法局部替换法团背装箱限制法局部替换法MAX SAT最大可满足性MAX-SAT(MAX-SAT): 任给关于变元x,x2,…,x n的简单最大可满足性(MAX SAT):1析取式C,C2,…,C m 及正整数K, 问存在关于变元x1,x2,…,x n1,C2,…,C m中至少有K 个为真吗?的赋值使得CK1设判定问题Π=<D,Y>, Π′=<D′,Y′>, 如果D′⊆D, Y′=D′∩Y, 则Π′是Π的特殊情况, 称作Π的子问题.: 如果已知Π的某个子问题Π是NP难的, 则Π也是限制法:′NP难的⎯⎯只需把Π′的实例I 看作Π特殊情况的实例.定理MAX-SAT是NP完全的.证MAX-SAT∈NP. 要证SAT≤MAX-SAT. 任给SAT的实例pI, 对应的MAX-SAT的实例f(I): 其他不变，仅需令K=m.I MAX SAT)其他不变仅需令3SAT3元合取范式: 每一个简单析取式恰好有3个文字的合取范式.每个简单析取式恰好有三元可满足性(3SAT): 任给一个3元合取范式F, 问F是可满足的吗?定理3SAT NP.是完全的’证显然3SAT∈NP.要证SAT ≤p3SAT. 任给个合取范式F, 要构造对应的3元合SAT3SAT.任给一个合取范式,取范式F ′=f(F), 使得F是可满足的当且仅当F′是可满足的.设F C1∧C2∧…∧C m, 对应的F F1∧F2∧…∧F m, F j是对=,′=′∧′′,′的合取范式, 并且应CjC j是可满足的当且仅当F j是可满足的.′F j′的构造规则(1) C j = z1. 引入两个新变元y j1, y j2, 令(1)F j′= (z1∨y j1∨y j2)∧(z1∨¬y j1∨y j2)∧(z1∨y j1∨¬y j2)∧(z1∨¬y j1∨¬y j2)(2)引入个新变元(2) C j = z1∨z2. 引入一个新变元y j, 令F j′= (z1∨z2∨y j)∧(z1∨z2∨¬y j)(3) C j = z1∨z2∨z3. 令F j′=C j.(3)(4) C j = z1∨z2∨…∨z k, k≥4. 引入k-3个新变元y j1, y j2,…,y j(k-3), 令F j′= (z1∨z2∨y j1)∧(¬y j1∨z3∨y j2)∧(¬y j2∨z4∨y j3)∧…∧(¬y j(k-4)∨z k-2∨y j(k-3))∧(¬y j(k-3)∨z k-1∨z k)实例:C j = z1∨z2∨…∨z5jF j′= (z1∨z2∨y j1)∧(¬y j1∨z3∨y j2)∧(¬y j2∨z4∨z5))1)1证明t(C)=1 ⇔t(F j′)=1j, 则存在i使得t(z i)=1设赋值)1t 满足Cj)=0 (1≤s≤k-3)当i=1 或2 时, 令t(yjs)=1 (1≤s≤k-3)当i= k-1 或k 时, 令t(yjs)=1 (1≤s≤i-2), t(y js)=0 (i-1≤s≤k-3)当3≤i≤k-2时, 令t(yjs′)=1.则有t(Fj反之, 设t(F′)=1.j)=0, 则t(z1∨z2)=1若t(yj1)=1, 则t(z k-1∨z k)=1若t(y)=1j(k-3)否则, 必有s(1≤s≤k-4) 使得t(y)=1且t(y j(s+1))=0,js)=1. 总之, 都有t(C j)=1.)1)1从而t(zs+2变换时间F 4j ′中简单析取式的个数不超过C j 中文字个数的4 倍, 每个简单析取式有3个文字, 因此可以在|F | 的多项式时间内构造出F ′.证明方法局部替换法证明方法：局部替换法要证Π1 ≤p Π2. 当Π2 是Π1 的子问题或两者的结构相似时, 往把实每个结构替换实往可以把Π1 的实例的每一个子结构替换成对应的Π2 实例的子结构.例如把SAT 的每个析取式C j ，替换成一组析取式F j ′6顶点覆盖团与独立集顶点覆盖、团与独立集引理对任意无向图G=<V,E>和子集V′⊆V, 下述命题等价: (1) V′是G 的顶点覆盖,集(2) V-V′是G 的独立集,(3) V-V′是补图G c = <V,E c> 的团.顶点覆盖(VC): 任给一个无向图G = <V,E>和非负整数K≤|V|,G K问G 有顶点数不超过K 的顶点覆盖吗?团: 任给一个无向图G = <V,E>和非负整数J ≤|V |, 问G 有顶点数不小于J 的团吗?J独立集: 任给一个无向图G = <V,E>和非负整数J ≤|V |, 问G 有顶点数不小于J 的独立集吗?J顶点覆盖定理顶点覆盖是NP 完全的.证:VC 的非确定型多项式时间算法: 任意猜想一个子集V ′式⊆V , |V ′|≤K , 检查V ′是否是一个顶点覆盖.3SAT ≤VC. x x …x n 3要证p 任给变元1,2,,n 的元合取范式F = C 1∧C 2∧…∧C m ,j =,i i .其中C j z j 1∨z j 2∨z j 3, z jk 是某个x i 或¬x i .如下构造VC 的实例f (F ): G = <V ,E > 和K =n +2m , 其中=,V V 1∪V 2, E =E 1∪E 2∪E 3,{}变换实例U = {x 1, x 2, x 3, x 4}, C =(x 1∨¬x 3∨¬x 4)∧(¬x 1∨x 2∨¬x 4)K = 4 + 2×2 = 8x 1x 1x 2x 2x 3x 3x 4x 4[x 2,2][x 3,1][x 4,2][x 1,2][x 1,1][x 4,1]10说明定理独立集和团是NP完全的.构件设计法顶点覆盖问题的NP完全性证明中设计了2种“构件”变元构件：真值赋值功能简单析取式构件：满足性检验功能,式用这些构件及构件之间的连接构成G, 通过这种方式达到用VC的实例表达3SAT的实例的目的哈密顿回路与货郎问题有向哈密顿回路: 任给有向图D , 问:D 中有哈密顿回路吗?定理有向HC 是NP 完全的.证要证3SAT ≤p 有向HC. 任给变元x 1,x 2,…,x n 的3元合取范式F =C ∧C ∧…∧C , C =∨∨, x i ¬x i .12m ,其中j z j 1z j 2z j 3,每个z jk 是某个i 或i 采用构件设计法构造有向图D . 表示变元x i 的构件是一条由串水平的顶点组成的链相邻的两个顶点之间有对方一串水平的顶点组成的链L i , 相邻的两个顶点之间有一对方向相反的有向边. 只有两种可能的方式通过L i 上的所有顶点⎯⎯从左到右或者从右到左通过L i 上的所有顶点, 这恰好对应x i 的值为1或者为0. 表示简单析取式C j 的构件是一个顶点c j .添加s 0,s 1,…,x n , 并通过它们把L 1,L 2,…,L n 连接起来.变元构件与析取式构件连络边链1L i 有3m +1的顶点, 依次为d i 0, a i 1, b i 1,d i 1,a i 2, b i 2,d i 2, …,a im , b im , d im . 对每一个C j =z j 1∨z j 2∨z j 3,(1)如果>>z jk =x i , 则添加<a ij ,c j > 和<c j ,b ij >; (2)如果z jk =¬x i , 则添加<c j ,a ij > 和<b ij ,c j >.实例C 2= x 1∨¬x 3∨¬x 4 对应的连接变换实例:x, x2, x3, F=(x1∨¬x2)∧(x2∨¬x3)1t(x1)=1, t(x2)=1, t(x3)=0)1)1)0s0C1C2s3证明F 证明：F 可满足⇒D 存在HC 设构造条从t 是F 的成真赋值. 要根据t 构造一条从s 0到s n, 最后回到s 0的哈密顿回路.(1) 依次对i =1,2,…,n 进行, 若t (x i )=1, 则从s i -1到d i 0, 从左到右经过L i 的所有顶点到达d im , 再到s i ; 若t (x i )=0, 则从s i -1到d im , 从右到左经过L i 的所有顶点到达d i 0, 再到s i . 最后, 从s n 回到s 0.(2) 将所有c j 插入这条回路. 设C j =z j 1∨z j 2∨z j 3, 由于t (C j )=1, 必有k (1≤k ≤3)使得t (z jk )=1. 若z jk =x i , 则通路从左到右经过L i , 且有有向边<a ij ,c j >和<c j ,b ij >. 于是, 可以把c j 插在a ij 与b ij 之间; 若z jk =¬x i , 则通路从右到左经过L i , 且有有向边<b ij ,c j >和<c j ,a ij >. 于是, 可以把c j 插在b ij 与a ij 之间. 这就得到D 中的一条哈密顿回路.证明F证明：D存在HC⇒F 可满足反之, 设D有哈密顿回路P, 不妨设P从s0到sn再到s结束. 正常的回路从左到右或从右到左通过每一条Li, 每个c j 插在某个a ij和bij 或者bij和aij之间. 若P从左到右通过Li, 令t(x i)=1; 若P从右到左通过Li, 令t(x i)=0. 不难证明必有t(C j)=1.要证P一定是正常的. 假设不然, P一定从某条链Ls 的顶点u到cj后没回到L而是到链L s≠t). u=a, 由于b只与a c及d相s t()若sj,sj sj,j sj邻, P已经过asj 和cj, b sj只剩下一个相邻的顶点, 故P不可能通b. u=b, P a. P过sj若sj,同理可证不可能通过sj都与是哈密顿回路矛盾, 所以P一定是正常的.构造D可以在多项式时间内完成.HC与TSP定理HC是NP完全的.HC. 任给一个有向图D=<V,E>, 要构造无证要证有向HC ≤p向图G=<V′,E′>使D有哈密顿回路当且仅当G有哈密顿回路.把D的每一个顶点v替换成3个顶点v in, v mid和v out, 用边连接v in 和v mid, v mid和v out. D的每条有向边<u,v>在G中换成(u out,v in).即V′={v in, v mid, v out| v∈V },E′={(u out,v in) | <u,v>∈E}∪{(v in,v mid),(v mid,v out) | v∈V }.定理TSP是NP完全的.恰好覆盖恰好覆盖:,a2,…,a n}，A的子集的集合W={S1,S2,…,S m}, 有穷集A={a1问: 存在子集U⊆W使得U中的子集都是不相交的且它们的并集中集交集等于A? 称W这样的子集U是A的恰好覆盖.例如, 设A={1,2,3,4,5}, S={1,2}, S2={1,3,4}, S3={2,4}, S4={2,5},1则{S2,S4}是A的恰好覆盖. 若把S4改为S4{3,5}, 则不存在A的.={3,5},恰好覆盖.定理恰好覆盖是NP完全的.构件设计∧∧证任给变元x 1,x 2,…,x n 的合取范式F =C 1∧C 2∧…∧C m , 其中A ={{|1t 1j m }jjs j j j z z z C ∨∨∨= 21A = { x 1,x 2,…,x n , C 1,C 2,…,C m }∪{ p jt | 1≤t ≤s j , 1 ≤j ≤m },其中p jt 代表C j 中的文字z jt . W 包含下述子集:变元构件恰好覆盖对每个：恰好覆盖对每个i ，T T i 与T F i 恰好取其1T T i ={x i }∪{p jt | z jt = ¬x i , 1≤t ≤s j , 1≤j ≤m }, 1≤i ≤n ,T F i ={x i }∪{ p jt | z jt = x i , 1≤t ≤s j , 1≤j ≤m }, 1≤i ≤n ,析取式构件：恰好覆盖对每个j ，C jt 中恰好取其1j C jt ={C j ,p jt }, 1≤t ≤s j , 1≤j ≤m .填充物{p jt }, 1≤t ≤s j , 1≤j ≤m.实例1F=(x1∨¬x2)∧(¬x1∨x2∨x3)∧(x1∨¬x2∨x3)∧(x1∨¬x3)A={x1,x2,x3,C1,C2,C3,C4, p11,p12,p21,p22, p23,p31,p32,p33,p41,p42}, W包含下述子集:{p11},{p12},{p21},{p22},{p23},{p31},{p32},{p33},{p41},{p42},T1T={ x1,p21}, T1F={ x1,p11, p31, p41},{{}T2T ={ x2,p12, p32}, T2F ={ x2,p22},T{F{}T3T ={ x3,p42}, T3F ={ x3,p23,p33},C11={C1, p11}, C12={C1, p12},C21={C2, p21}, C22={C2, p22}, C23={C2, p23}, {}{}{}C31={C3, p31}, C32={C3, p32}, C33={C3, p33},C41={C4, p41}, C42={C4, p42}{}{23恰好覆盖满设证明：U 是恰好覆盖⇔F 可满足U ⊆W 是A 的恰好覆盖, ∀i , 若T T i ∈U , 则令t (x i )=1;若T F i ∈U , 则令t (x i )=0. ∀j , 必有一个C jt ={C j ,p jt }∈U . z jt =x i 或¬x i .(1))1(1) 若T T i ∈U , 则p jt ∉T T i , 从而z jt =x i . 有t (x i )=1, 故t 满足C j .(2) 若T F i ∈U , 则p jt ∉T F i , 从而z jt =¬x i . 有t (x i )=0, 故t 也满足C j . F t 是F 的成真赋值.反之,t .,)=1,T ;)=0,, 设t 是F 的成真赋值. ∀i , 若t (x i )1, 则U 包含T i ; 若t (x i )0, 则U 包含T F i . ∀j , 由于t 满足C j , C j 必有一个文字z jt 使得t (z jt )=1, .,.,从而U 中现有的子集不包含p jt . 于是, 可以把C jt 加入U . 至此, U 覆盖了所有的x i 和C j , 以及部分p jt . 最后, 把尚未被覆盖的p jt 构成的单元子集{p jt }加入U , 即可得到A 的恰好覆盖.24变换时间由于F 中的文字数不超过mn, 故|A| ≤n + m + mnW 中的子集数不超过2n + 2mn每个子集的大小不超过n+1.而且构造很简单, 显然可以在多项式时间内完成.子集和,背包,装箱与双机调度子集和: 给定正整数集合X ={x 1,x 2,…,x n }及正整数N , 问存在X 的子集T , 使得T 中的元素之和等于N 吗? 装箱: 给定n 件物品, 物品j 的重量为正整数w j , 1≤j ≤n , 以及箱K . B , 子数规定每只箱子装入物品的总重量不超过正整数,问能用K 只箱子装入所有的物品吗?双机调度: 有2台机器和n 项作业J 1,J 2,…,J n , 这2台机器完全相同, 每一项作业可以在任一台机器上进行, 没有先后顺序, 1作业J i 的处理时间为t i , 1≤i ≤n , 截止时间为D , 所有t i 和D 都是正整数, 问能把n 项作业分配给这2台机器, 在截止时间D 内完成所有的作业吗?子问题关系子集和是0-1背包的子问题= v i且B = K.限制0-1背包的实例中所有wi双机调度是装箱问题的子问题物品看作作业, 物品的重量是作业的处理时间, 截止时间是每只箱子允许的最大重量.限制箱子数K=2只需证明子集和与双机调度问题的NP完全性构件设计定理子集和是NP完全的.,a2,…,a n}和W={S1,S2,…,S m},证恰好覆盖：有穷集A={a1子集和：非负整数集合X={x,x2,…,x n}及非负整数N.1和N都可表成kn位的二进制数, 这kn位分成n 段, 每每个xj(m+1)⎤. N的每一段的第一位(最右的一位)为1, 段k位, k=⎡log2其余的为0. x对应于子集S. 当a i∈S时, 从左到右x的第i段j j j j的第一位为1, 其余的为0.变换实例n=4, m=3, A={a1,a2,a3,a4}, S1={a1,a2}, S2={a1,a3,a4}, S3={a2}, N=01010101, x1=01010000, x2=01000101, x3=00010000.=01010101=01010000=01000101=00010000存在集和W含A的恰好覆盖⇔存在子集和=N| S j∈U}. 由于A中的每个元U⊆W是A的恰好覆盖, 令T={x设|}中的每一个元j中恰好出现一次, 故对于二进制数的每一段, 素在U的所有Sj中恰好有一个的这一段为00…01, 从而T 中所有在T的所有x01Tj元素之和等于N.反过来, 设X的子集T 中元素之和等于N, 令U={S| x j∈T}. T 中j(m+1)⎤位, 最大值为2k-1 ≥m, 至多有m个数, 每一段有k=⎡log2故T 中的数相加时不会出现段之间的进位. 对于每一段, 在T的所有x中恰好有一个的这一段为00…01, 这意味着每一个a ji 中恰好出现一次, 即U是A的恰好覆盖.在U的所有Sj构造X和N显然可以在多项式时间内完成,010-1背包与伪多项式时间算法定理0-1背包是NP完全的.注意: 0-1背包问题优化形式的动态规划算法, 其时间复杂度意算法度为O(nB), 其中n是物品的个数, B是重量限制. 这不是多项式时间算法, 而是指数时间算法.伪多项式时间算法: 算法的时间复杂度以|I|和max(I)的某个二元多项式p(|I|, max(I))为上界, 其中max(I)是实例I中数的最大绝对值.双机调度与装箱定理双机调度是机调度是NP完全的. 证要证子集和≤p双机调度. 任给子集和实例:非负整数集X={x1, x2,…, xn}及非负整数N, 对应的双机调度实例有n+2项作业J1, J2,…, Jn+2, 处理时间为 x1, x2,…, xn, a, b, 截止时间为D. 要求存在 X 的子集T 使得当且仅当 N+a = M-N+b=D. 于是, a = -2N+b. 取 b=M +2N, a=2 2M , D=2 2M+N, 其中 M= x1+x2+…+ xn.变换实例子集和实例 X={ 5, 2, 7, 4, 3 }, N=6 子集 T={ 2,4 2 4 }, } 2+4=6 M=21, a=2M=42, b=M+2N=33, D=2M+N=48 双机调度实例 J ={ { J1, J2, J3, J4, J5, J6, J7 } t ={ 5, 2, 7, 4, 3, 42, 33 } D=48 可行调度 J2+J4+J6: 2+4+42=48；J1+J3+J5+J7: 5+7+3+33=4832NPC证明小结证明问题是NP完全的 (1) 证明该问题属于NP. (2) 选好一个已知选好个已知NP完全问题变换到该问题, 给出变换规则. (3) 证明肯定实例当且仅当变换到肯定实例. (4) 证明变换时间是多项式的——变换规模是多项式规模. 常用的证明方法限制法局部替换法构件设计法NP完全性理论的应用用NP完全性理论进行子问题分析完全性论进行子问题分析子问题的计算复杂性子问题的NP完全性证明搜索问题 Turing归约 NP-hard, NP-easyNP难度 34子问题的计算复杂性NPC?P 努力扩大已知区域，缩小未知区域当P≠NP时，存在不属于NPC也不属于P的问题35有先行约束的多处理机调度问题优化问题: 给定任务集 T, m 台机器,∀t∈T, l(t)∈Z+, T上的偏序‫ط‬. 若σ:T→{ 0, 0 1 1, … , D }满足下述条件,则称 T 为可行调度. (1) ∀t∈T, σ(t)+l(t)≤D (2) ∀i, 0 ≤ i≤ D, | { t∈T : σ(t) ≤ i < σ(t)+l(t) }| ≤ m (3) ∀t, t’∈T, t ‫ ط‬t’⇔ σ(t)+l(t) ≤ σ(t’) 求使得 D 最小的可行调度. 条件说明：任务在截止时间前完成同时工作的台数不超过 m 有偏序约束的任务必须按照约束先行36实例任务集如图 m=2, 求使得 D 最小的可行调度.t611 1 1t522t4t3t2t1t6 t5 t4 t6 t5 t3t4t2 ∅ t2 t3 ∅ t1t1 D=6 6D=537对应的判定问题实例：任务集实例任务集T, m台机器,∀t∈T, l(t)∈Z+,T上的偏序‫ط‬, 截止时间 D∈Z+. 问是否存在小于等于 D 的可行调度？问：是否存在小于等于的行调度子问题通过限制参数(机器数、工作时间、偏序)构成参数偏序 m大小 l 大小 ∅ l 为常数限制树任意 m 任意 l 任意38m ≤1, 2, … , 某个常数调度问题的子问题结构从上到下：偏序任意、树形偏序、无偏序约束从左到右：处理器台数限制逐步放大从前到后：各任务等长工作时间、任意工作时间l 任意 ≺ l=1任意为树≺≺ =∅m=1m≤2m≤3m≤Mm任意39某些子问题属于Pm =1, 1 l 任意, 任意偏序任意算法： 1 计算所有任务的工作时间之和 Time 1. i 2. if Time ≤ D,存在可行调度. 3 按照偏序从高层依次取任务，进行安排即可 3. 按照偏序从高层依次取任务进行安排即可.m 任意, l 为常数, 偏序为空算法： 1. Time ←⎡|T|/m⎤ l 2. if Time ≤ D,存在可行调度. 3. 将任务按顺序依次分配到 m 台机器上.40某些子问题属于Pm任意，l 为常数，偏序为树关键路径算法：1.按照偏序关系，从树叶依次拿任务，每次至多下移m 个结点.2.将分配次数与D比较,如果不超过D, 回答Yes.分配次数至少为树高,可以证明时间为多项式时间.m=2, 偏序任意, l 为常数算法：H.N.Gabow,An almost linear algorithmfor two processors scheduling, J.Assoc.Comput.Math,29, 1982, 766-780.41术语简介极大多项式可解（加绿圈）的子问题π：（加绿圈）π是多项式时间可解的,并且不存在其他多项式’’的子问题可解的子问题π,使得π是π的子问题.极小NP完全的子问题π（加篮圈）π是NP完全的，并且不存在其他NP完全的子问题π’使得π’是π的子问题.极小未解决的子问题π：（加红圈）π的子问题都属于P, 但不知π是否属于P，也不知P是否属于NPC.极大未解决的子问题π: (加粉圈)π的父问题都是NP完全的，但不知π是否属于P，也不知P是否属于NPC.42NP完全子问题的证明许多子问题的NP完全性证明采用局部替换法.以图论为例，平面性限制，顶点度数限制等形成子问题.以顶点度数D 为例，子问题分析如下：问题P NPCVC D≤2D≥3HC23顶点三着色34反馈顶点集23团给定D任意43Turing 归约多项式时间的Turing 归约：设π, π是搜索或优化问题，A 是利用解π2的假想子程序s 12 2 解π1的算法，且只要s 是多项式时间的，A 也是多项式时间2.的，则称算法A 是从π1到π2 的多项式时间的Turing 归约. 这是也称π1 Turing 归约到π2，记作π1∝T π2.判定问题是搜索问题特例，多项式变换等价于Turing 归约. 对于判定问题π，若I 属于Y π, 则S π[I ]={ Yes }, 若I 属于D −Y , 则S I ]={ No }.44ππ,π[]{}图灵归约的性质传递性π1∝T π2，π2∝T π3⇒π1∝T π3多项式变换是Turing 归约的特例.设π多项式变换到π, 如下设计解π1的算法A ：12, 1 1. 对于π1的任何实例I ，先将I 变换成π2实例I’,22I’2. 利用解π2 的假想子程序s 对I 识别.3. 如果I’是肯定实例，则I 也是肯定实例；4是否定实例则I 4. 如果I’是否定实例，则I 也是否定实例.易见只要s 是多项式时间的，则A 也是多项式时间的. 45因此π1∝T π2.T iTuring归约的应用证明许多NPC 问题所对应的优化或构造问题为NP-hard. 证明某些非NP 类的判定问题是NP-hard说明:将解对应优化问题或构造问题的算法作为解判定问题算法的子程序，得到解判定问题的算法，从而构造了从判定问题到对应优化问题或构造问题的Turing归约.证明某些NPhard问题是NP等价的46NP等价证明：货郎问题等价证明货郎问题例1 货郎问题（TSO）是NP等价的.证：易证TSO是NP-hard. 下面证明TSO 是NP-easy.引入中间问题：货郎问题的延伸问题（TSE）TSE实有市集实例：有穷城市的集合C= { c, c2, … , c m}1, c j∈C, d(c i, c j)∈Z+,距离∀ci长度限制B∈Z+,部分旅行路线ϑ= < c, … ,cπ(k) >π(1)问：ϑ是否可以延伸成全长不超过B的全程旅行问< cπ(1), … , cπ(k), cπ(k+1), … , cπ(m)>?易证TSE属于NP.NP47TSO 到TSE 的图灵归约设s (C ,d ,ϑ,B )是解TSE 的子程序，其中C 为城市集，d 为距离函数，ϑ为部分旅行，B 为长度限制.下面构造解TSO 的算法.思路：用二分法确定最短路旅行长度B *旅行长度界于m →m ×d , d = max{ d (c i , c j )}每次取中点值验证是否存在能延伸到此长度的旅行根据最小长度值B *确定旅行路线开始依次检查><>从c 1开始，依次检查<c 1,c 2>,<c 1,c 3>…是否能延伸到B *长度的旅行，选择第一个可延伸的顶点c i . .48按照上面方法确定后面的其他顶点.算法MinLength设S的子程序其中s(C,d,ϑ,B)是解TSE 的子程序，其中C为城市集，d为距离函数，ϑ为部分旅行，B为长度限制.算法Minlength（二分法确定最短旅行长度B*）令min,max{(i,j)i,j}1. B←m, B←m⋅max{ d c c): c, c∈C2. 若B max−B min=1, 则B*←B max, 结束3.i3. B←[(B min+B max)/2]4. s(C,d,<c1>,B)5. 若回答Yes”, B max←B, 否则B min←B5若回答”Yes”6．转2.49算法图示B min =m B max =md B=(B min +B max )/2Yes B min =m B max =B i =B =md No B min B B max md 50B min B*=B max return。

lecture 10

期权
• 期权就是事先以较小的代价购买一种在未来规定的时间内以某一确定价格买入或卖出某种金融工具的权利。其中，购买这种权利所费的代价就是权利金，而未来买入或卖出某种金融工具时的价格就是履约价格。
按照履约时间的不同，期权分为：
• 美式期权：期权的购买者可以在期权到期日以及到期日之前的任何时间里执行权利的期权。 • 欧式期权：只能在期权的到期日执行期权。
买入期权举例
• 假设益智公司股票当前的价格为15元，你预计它在六个月后会上涨到22元，但你又不敢肯定。于是，你以每股1元的价格从王小二手里购买了这样一份买入期权，约定在六个月后你有权利从王小二手里以每股17元的价格购买1万股益智股票的权利。如果六个月后股票的价格真的上涨到了22元，那么，你就执行以每股17元的价格从王小二手里买入 1万股益智股票的权利，然后以每股22元的价格将其在现货市场上卖掉。
买入价 4550 现货市场 4800 买入价 4500
Currency options
• Currency options give the holder the right to exercise the option until its expiration. • American style options & European style options • the option seller or writer & the option buyer
期货与期货交易
•期货合约就是在期货交易所内达成的、标准化的远期合同。 •期货交易指的是买卖双方事先就交易的商品数量、质量等级、交割日期、交易价格和交割地点等达成协议，而在约定的未来某一日期进行实际交割的交易。

lecture 10

Techniques of Molecular Biology1.Electrophoresis2.Restriction digestion3.Hybridization4.PCR5.Genome sequence & analysis6.DNA Cloning and gene expressionGel electrophoresis separates DNA and RNA molecules according to size shape and topological propertiesTo separate DNA of different size rangesNarrow size range of DNA: use polyacrylamideWide size range of DNA: use agarosegelVery large DNA(>30-50kb): use pulsed-field gel electrophoresisSouthern/Northern/Western Blotting, those techniques for what’s purpose ? Southern Blotting detects the target gene in genomeNorthern Blotting detects the RNA（usually mRNA）and expression level Western Blotting to detect specific proteinPCR cycle principles①Denaturation: The target DNA (template) is separated into two stands by heating to 95℃②Primer annealing: The temperature is reduced to around 55℃ to allow the primers to anneal.③Polymerization (elongation, extension):The temperature is increased to 72℃ for optimal polymerization step which uses up dNTPsand required Mg++.The requirement of PCR cycle①Template ：PCR can only be applied ifsome sequence information is known so that primers can be designed.②Primers：PCR primers need to be about 18 to 30 nt long and have similar G+C contents so that they anneal to their complementary sequences at similar temperatures.They are designed to anneal on opposite strands of the target sequence. Way to PCR optimizationWe can change the annealing temperature and the Mg++ concentration or carry out nested PCR to optimize PCR.Two ways for sequencing:1. DNA molecules (radioactivellabeled at 5’termini) are subjected to 4 regiments to be broken preferentially at Gs, CsTs, As, separately.2. chain-termination method。

Lecture_10

14-6
Open Market Purchase from Nonbank Public II
Nonbank Public Federal Reserve System
Assets
Securities Currency -$100 +$100
Liabilities
Assets
Securities
Liabilities
(borrowing from Fed)
(borrowing from Fed)
• Monetary liabilities of the Fed have increased by $100 • Monetary base also increases by this amount
Copyright © 2010 Pearson Addison-Wesley. All rights reserved.
14-11
Paying Off a Discount Loan from the Fed
Banking System
Assets Reserves Liabilities -$100 Discount loans -$100
Federal Reserve System
Assets Discount loans Liabilities -$100 Reserves -$100
14-8
Open Market Sale
Nonbank Public Federal Reserve System
Assets
Securities Currency +$100 -$100
Liabilities
Assets
Securities

Lecture10_序列密码

密钥k 初始向量IV 密钥流生成器密钥流生成器密钥流 ki 明文流mi 密文流ci 加密算法E 解密算法D 密钥流 ki 明文流mi
图10-3 自同步序列密码模型

特点：
– 自同步
– 有限的错误传播
4
5
序列密码和分组密码

序列密码的优点
– 转换速度快 – 低错误传播

序列密码的缺点
LFSR 1
AND 时钟 LFSR 1
LFSR 2
LFSR 2
＋
x3 AND g(x)
密钥流
＋
AND LFSR 3
输出
LFSR 3
图10－8 Geffe生成器
图10－9 交错停走式生成器
14
利用LFSR的序列密码反馈加密体制
明文 m
mi
ci
＋
ki
＋
＋
...
＋
AND 密文 c
（4）其结构能够应用代数方法进行很好的分析
10
an
an-1
an-2
a2
a1
f(a1,a2,…,an)
图10-5 反馈移位寄器
11
＋ an+1 AND
＋
...
＋
AND
AND
Cn-1
AND
c1
an
c2
an-1 ...
cn
a1 输出
a2
图10－6 长度为n的线性反馈移位寄存器
12
例子
＋ a5 AND AND
18
加密与解密

加密
– 将K与下一明文字节异或

解密
– 将K与下一密文字节异或

Lecture 10 词语的搭配、语法形式与翻译

3) Thomas Edison surround himself with great thinkers and was friend with some the most successful people of his day. He would have valued many of their ideas, some very similar to his own and others in contrast to his thinking. 爱迪生的周围聚集了许多伟大的思想家，他的朋友包括了当时一些最成功的人士。他们有些想法与爱迪生的想法不谋而合，有的却大异其趣，但他们的许多想法或许都曾得到了爱迪生的重视。

1) The second aspect is the application by all members of society, from the government official to the ordinary citizen, of the special methods of thought and action that scientists use in their work. 第二个方面是全体社会成员——从政府官员到普通公民——都要使用科学家在他们的工作中所使用的那些特殊的思考和行动的方法。注：关键搭配——application of …

2) During the Dark Ages, anyone who believed in science would be burned as a heretic. 在中世纪，谁要是相信科学就会被视为异教徒而被烧死。注：关系从句暗含假设意义。

lecture10

Lecture 10Analysis of Gene SequencesAnatomy of a bacterial gene:S-D SequenceSequence Element FunctionPromoter To target RNA polymerase to DNA and to start transcriptionof a mRNA copy of the gene sequence.Transcription terminator To instruct RNA polymerase to stop transcription.Shine-Dalgarno sequence and translation start S-D sequence in mRNA will load ribosomes to begin transla-tion. Translation almost always begins at an AUG codon in the mRNA (an ATG in the DNA becomes an AUG in the mRNA copy). Synthesis of the protein thus begins with a methionine.Coding Sequence Once translation starts, the coding sequence is translated bythe ribosome along with tRNAs which read three bases at atime in linear sequence. Amino acids will be incorporated intothe growing polypeptide chain according to the genetic code. Translation Stop When one of the three stop codons [UAG (amber), UAA(ochre), or UGA] is encountered during translation, thepolypeptide will be released from the ribosome.Example: A gene coding sequence that is 1,200 nucleotide base pairs in length (including the ATG but not including the stop codon) will specify the sequence of a protein 1200/3 = 400 amino acids long. Since the average molecular weight of an amino acid is 110 da, this gene encodes a protein of about 44 kd — the size of an average protein.The Genetic CodeClassically, genes are identified by their function. That is the existence of the gene is recognized because of mutations in the gene that give an observable phenotypic change. Historically, many genes have been discovered because of their effects on phenotype. Now, in the era of genomic sequencing, many genes of no known function can be detected by looking for patterns in DNA sequences. The simplest method which works for bacterial and phage genes (but not for most eukaryotic genes as we will see later) is to look for stretches of sequence that lack stop codons. These are known as “open reading frames” or ORF s. This works because a random sequence should contain an average of one stop codon in every 21 codons. Thus, the probability of a random occurrence of evena short open reading frame of say 100 codons without a stop codon is very small (61/64)100 = 8.2 x 10–3Identifying genes in DNA sequences from higher organisms is usally more difficult than in bacteria. This is because in humans, for example, gene coding sequences are separated by long sequences that do not code for proteins. Moreover, genes of higher eukaryotes are interrupted by introns, which are sequences that are spliced out of the RNA before translation. The presence of introns breaks up the open reading frames into short segments making them much harder to distinguish from non-coding sequences. The maps below show 50 kbp segments of DNA from yeast, Drosophila, and humans. The dark grey boxes represent coding sequences and the light grey boxes represent introns. The boxes above the line are transcribed to the right ant the boxes below are transcribed to the left. Names have been assigned to each of the identified genes. Although the yeast genes are much like those of bacteria (few introns and packed closely together), the Drosophila and human genes are spread apart and interrupted by many introns. Sophisti-cated computer algorithms were used to identify these dispersed gene sequences. Saccharomyces cerevisiaeYFL046W YFL040W YFL030WYFL044C YPT1 RPL22B CAF16YFL042C GYP8Drosophila melanogasterCG3131To see how gene sequences are actually obtained, we will first need to consider somefundamentals of the chemical structure of DNA. Each strand of DNA is directional. The different ends are usually called the 5’ and 3’ ends; referring to different positions on the ribose sugar ring where the linking phosphate residues attach.ded DNA molecule the two strands run anti-parthe general structure can be diagramed like this: ’’35 5’3’• Note about representation of DNA sequences.1) Single strands are always represented in direction of synthesis – 5’ to 3’ 2) For double stranded DNA, usually one strand is represented in the 5’ to 3’ direction. For a gene, the strand represented would correspond to the sequence of the mRNA. DNA polymersaes are the key players in the methods that we will be considering. The general reaction carried out by DNA polymerase is to synthesize a copy of a DNA template starting with the chemical precursors (nucleotides) dATP, dGTP, dCTP, and dTTP (dNTPs). All DNA polymerases have two fundamental properties in common.(1) New DNA is synthesized only by elongation of an existing strand at its 3’ end.(2) Synthesis requires nucleotide precursors, a free 3’ OH end, and a template strand.A general substrate for DNA polymerase looks like this:’’ 3 ’553’ Note that the template strand can be as short as 1 base or as long as several thousand bases.After addition of DNA polymerase and nucleotide precursors this product will be readily synthesized: ’’35 5’3’DNA SequencingConsider a segment of DNA that is about 1000 base pairs long that we wish to sequence.(1) The two DNA strands are separated. Heating to 100˚C to melt the base pairing hydrogen bonds that hold the strands together does this.(2) A short oligonucleotide (ca. 18 bases) designed to be complimentary to the end of one of the strands is allowed to anneal to the single stranded DNA. The resulting DNA hybrid looks much like the general polymerase substrate shown previously.(3) DNA polymerase is added along with the four nucleotide precursors (dATP, dGTP, dCTP, and dTTP). The mixture is then divided into four separate reactions and to each reaction a small quantity different dideoxy nucleotide precursor is added. Dideoxy nucleotide precursors are abbreviated ddATP, ddGTP, ddCTP, and ddTTP.(4) The polymerase reactions are allowed to proceed and, using one of a variety of methods, radiolabel is incorporated into the newly synthesized DNA.(5) After the DNA polymerase reactions are complete, the samples are melted and run ona gel system that allows DNA strands of different lengths to be resolved. The DNA sequence can be read from the gel by noting the positions of the radiolabeled fragments.The crucial element of the sequencing reactions is the added dideoxynuclotides. These molecules are identical to the normal nucleotide precursors in all respects except that they lack a hydroxyl group at their 3’ position (3’ OH).A part of the final gel will look like this:+ddGTP +ddATP +ddTTP +ddCTPTop (—)Bottom (+)(Note that larger molecules migrate more slowly to the cathode on these gels) The deduced DNA sequence obtained from this gel is: 5’ GGATCCTATC 3’Polymerase Chain ReactionNow let’s consider how to obtain DNA segments that are suitable for sequencing. At first, DNA sequences were obtained from cloned DNA segments (we will discuss some methods to clone new genes in a subsequent lecture). Presently the entire DNA sequence for E. coli, as well as a variety of other bacterial species, has been determined. If we want to find the sequence of a new mutant allele of a known gene we need an easy way to obtain a quantity of this DNA from a culture of bacterial cells. The best way to do this is to use a method known as PCR or polymerase chain reaction that was developed by Kary Mullis in the mid-1980’s. The steps in a PCR reaction are as follows.(1) A crude preparation of chromosomal DNA is extracted from the bacterial strain of interest.(2) Two short oligo nucleotide primers (each about 18 bases long) are added to the DNA. The primers are designed from the known genomic sequence to be complimentary to opposite strands of DNA and to flank the chromosomal segment of interest.(3) The double stranded DNA is melted by heating to 100˚C and then the mixture is cooled to allow the primers to anneal to the template DNA.(4) DNA polymerase and the four nucleotide precursors are added and the reaction is incubated at 37˚C for a period of time to allow a copy of the segment to be synthesized.(5) Steps 3 and 4 are repeated multiple times. To avoid the inconvenience of having to add new DNA polymerase in each cycle a special DNA polymerase that can withstand heating to 100˚C is used.The idea is that in each cycle of melting, annealing and DNA synthesis the amount of the DNA segment is doubled. This gives an exponential increase in the amount of the specific DNA as the cycles proceed. After 10 cycles the DNA is amplified 103 fold and after 20 cycles the DNA will be amplified 106 fold. Usually amplification is continued until all of the nucleotide precursors are incorporated into synthesized DNA.。

Lecture 10

• 第二，你初到一个餐馆，开始举筷时有新鲜感，新盖的茅房三天香，这也可以叫做 “陌生化效应”吧。(10)
Analysis: Attention should be paid to cohesive devices: reference, substitution, ellipsis, conjunction lexical cohesion
• Another example: • 为了推动中美关系的发展，中国需要进一步了解美国，美国也需要进一步了解中国。 • To promote the development of Sino-American relations, China needs to know the United States better, and vice versa. • S8:……饿了吃糠甜加蜜，饱了吃蜜也不甜. A: Even chaff tastes sweet as honey when one is hungry, whereas honey does not taste sweet at all when one is full. B: The least tasteful food taste sweet when you’re hungry while the sweetest honey tastes nothing special when you’re full. C: Husks will taste like honey when you are hungry, and vice versa. (better)
• S4: 开始吃头盘或冷碟的时候，印象很好。S5: 吃头两个主菜时，也赞不绝口。 A: When we begin with the first course or the appetizer, we are very positively impressed. When we continue with the first two main courses, we are also full of praises for the dishes. B: you will be not only favorably impressed with the first dish or cold dishes, but also profuse in praise of the first two courses. (better)

Lecture 10

从文体上看，广告标题从选词到句法都运用了英语的口语体，其创意在于扩大潜在的消费群体：睡眠与每个人都息息相关 “失眠”是日常生活中普遍存在的问题
同时，广告标题还借用简短句式所特有的于脆利落的文体效果，来传达这样的信息：解决失眠问题同样可以干脆利落如果您难以入睡，这不是什么难题我们的产品可以帮你调节睡眠广告撰写人正是通过这种朴实无华、不动声色的手法来激发人们的购买欲望。
Section two Public Service Advertisement, Commercial Advertisement and their Functions 公益广告和商业广告及其功能
一、公益广告和商业广告广告，按照其宣传目的，可以分为公益广告和商业广告两类。公益广告(public service advertisement)是不以盈利为目的、而为公众利益服务的广告。它意在唤起人们对共同的长远和重大利益的关注，强调人与人之间的互助行为，呼吁人们以实际行动来改善或解决各种社会问题。
口号是广告中的主题语，即用十分精练的语句概括出一则广告的主题，起到加深印象和宣传鼓动的作用，同时也可以体现商品和企业的形象。附文包括商标(trademark)、商品名称(name of commodity)、公司或厂家的视觉标记 (visual symbols)、公司地址、电话和商品价格等，旨在强化商品和企业形象，提供联系购买的信息。
This provides a naturallief of sleeplessness and insomnia, without causing the adverse side effects associated with some convenient sleeping remedies. So for a perfect natural solution to sleeplessness and insomnia, try Brauer Sleep and Insomnia Relief. Available in blister-pack tablets or a 20ml oral spray．

lecture10

幻灯片1I N T E R N A T I O N A L E C O N O M I C SL e c t u r e10幻灯片2M A I N C O N T E N T S----e c o n o m i c g r o w t h a n d i n t e r n a t i o n a l t r a d e●F a c t o r g r o w t h a n d p r o d u c t i o n p o s s i b i l i t y f r o n t i e r●T e c h n i c a l p r o g r e s s a n d t h e n a t i o n”s P P F●G r o w t h a n d t r a d e:t h e s m a l l-c o u n t r y c a s e●G r o w t h a n d t r a d e:t h e l a r g e-c o u n t r y c a s e●I m m i s e r i z i n g g r o w t h幻灯片3F a c t o r g r o w t h a n d p r o d u c t i o n p o s s i b i l i t y f r o n t i e r(e c o n o m i c g r o w t h)●A n i n c r e a s e i n t h e e n d o w m e n t o f o n e o r m o r e t h a n o n e f a c t o r s s u c h a s l a b o r a n dc a p i t a l o v e r t i m e c a u s e s t h e n a t i o n’s p r od u c t i o n f r o n t ie r t o s h if t o u t w a r d.t h e t y p e a n dd e g r e e o f t h e s h i f t d e p e n d o n t h e r a t e l a b o r a n d c a p i t a l g r o w s.●I f l a b o r a n d c a p i t a l g r o w a t t h e s a m e r a t e,t h e n a t i o n’s p r o d u c t i o n f r o n t i e r w i l l s h i f to u t e v e n l y i n a l l d i r e c t i o n s a t t h e r a t e o f f a c t o r g r o w t h.T h i s i s t h e b a l a n c e d g r o w t h.幻灯片4●I f o n l y o n e f a c t o r e n d o w m e n t(.e.g.L)g r o w s,t h e o u t p u t o f b o t h c o m m o d i t i e s g r o w.B u t t h e o u t p u t o f c o m m o d i t y i n t e n s i v e i n t h e g r o w i n g f a c t o r i n i t s p r o d u c t i o n g r o w sf a s t e r t h a n t h a t c o m m o d i t y n o t i n t e n s i v e i n t h e f a c t o r i n i t s p r o d u c t i o n.●N o t e t h a t w h e n o n l y o n e f a c t o r o f p r o d u c t i o n d o u b l e s,t h e m a x i m u m o u t p u t O fc o m m od i t y i n te n s i v e i n幻灯片5●t h e g r o w i n g f a c t o r d o s e n o t d o u b l e,f o r t h e c o m m o d i t y t o d o u b l e,b o t h o r a l l f a c t o r su s e d i n p r o d u c t i o n m u s t d o u b l e.●N o t e t h a t I f t h e t o t a l p o p u l a t i o n g r o w s a t t h e s a m e r a t e a s t h e e c o n o m i c g r o w t h o rf a c t o rg r o w t h,r e a l p e r c a p i t a i n c o m e a n d th e w e l f a r e o f t h e n a ti o n t e n d t o r e m a i nu n c h a n g e d.幻灯片6●I f o n l y l a b o r g r o w s(o r L g r o w s p r o p o r t i o n a t e l y m o r e t h a n K),K/L w i l l f a l l a n d s o w i l lt h e p r o d u c t i v i t y o f L,t h e r e t u r n s t o L,a n d r e a l p e r c a p i t a i n c o m e.I f,o n t h e o t h e rh a n d,o n l y t h e e n d o w m e n t o f K g r o w s(o r K g r o w s p r o p o r t i o n a t e l y m o r e t h a n L），K/L w i l l r i s e a n d s o w i l l t h e p r o d u c t i v i t y o f L,t h e r e t u r n s t o L,a n d r e a l p e r c a p i t ai n c o m e.幻灯片7T e c h n i c a l p r o g r e s s a n d t h e n a t i o n”s P P F(e c o n o m i c g r o w t h)●S e v e r a l e m p i r i c a l s t u d i e s h a v e i n d i c a t e d t h a t m o s t o f t h e i n c r e a s e i n r e a l p e r c a p i t ai n c o m e i n i n d u s t r i a l n a t i o n s i s d u e t o t e c h n i c a l p r o g r e s s a n d m u c h l e s s t o c a p i t a la c c u m u l a t i o n.●T e c h n i c a l p r o g r e s s i s u s u a l l y c l a s s i f i e d i n t o n e u t r a l,l a b o r s a v i n g,o r c a p i t a l s a v i n gt e c h n i c a l p r o g r e s s.幻灯片8●N e u t r a l t e c h n i c a l p r o g r e s s i n c r e a s e s t h e p r o d u c t i v i t y o f L a n d K i n t h e s a m ep r o p o r t i o n,s o t h a t K/L r e m a i n s t h e s a m e a f t e r t h e n e u t r a l t e c h n i c a l p r o g r e s s a s i t w a s b e f o r e a t u n c h a n g e d r e l a t i v e f a c t o r p r i c e s(w/r),t h e r e i s n o s u b s t i t u t i o n o f L f o r K i n p r o d u c t i o n.A l l t h a t h a p p e n s i s t h a t a g i v e n o u t p u t c a n n o w b e p r o d u c e d w i t h l e s s L a n d l e s s K.幻灯片9●L a b o r–s a v i n g t e c h n i c a l p r o g r e s s i n c r e a s e t h e p r o d u c t i v i t y o f K p r o p o r t i o n a t e l ym o r e t h a n t h e p r o d u c t i v i t y o f L.A s a r e s u l t,K i s s u b s t i t u t e d f o r L i n p r o d u c t i o n,t h a ti s,m o r e K i s u s e d p e r u n i t o f L.●N o t e t h a t a g i v e n o u t p u t c a n n o w b e p r o d u c e d w i t h f e w e r u n i t s o f L a n d K b u t w i t h ah i g h e r r a t i o o f K/L.幻灯片10●C a p i t a l–s a v i n g t e c h n i c a l p r o g r e s s i n c r e a s e s t h e p r o d u c t i v i t y o f L p r o p o r t i o n a t e l ym o r e t h a n t h e p r o d u c t i v i t y o f K.A s a r e s u l t,L i s s u b s t i t u t e d f o r K i n p r o d u c t i o n,t h a ti s,m o r e L i s u s e d p e r u n i t o f K.●N o t e t h a t a g i v e n o u t p u t c a n p r o d u c e d w i t h f e w e r u n i t s o f L a n d K b u t w i t h a h i g h e rr a t i o o f L/K.幻灯片11●a l l t y p e s o f t e c h n i c a l p r o g r e s s c a u s e t h e n a t i o n’s P P F t o s h i f t o u t w a r d,t h e t y p e a n dd e g r e e o f t h e s h i f t d e p e n d o n t h e t y p e a n d r a t e o f t e c h n i c a l p r o g r e s s.●i t m u s t b e p o i n t e d o u t t h a t,i n t h e a b s e n c e o f t r a d e,a l l t y p e s o f t e c h n i c a l p r o g r e s si n c r e a s e a n a t i o n’s w e l f a r e幻灯片12G r o w t h a n d t r a d e:t h e s m a l l-c o u n t r y c a s e●T h e s m a l l-c o u n t r y c a s e m e a n s t h e n a t i o n i s t o o s m a l l t o a f f e c t t h e r e l a t i v e c o m m o d i t yp r i c e s a t w h i c h i n t e r n a t i o n a l t r a d e t a k e s p l a c e.●i f t h e f a c t o r o f p r o d u c t i o n r i s e s,i n w h i c h a n a t i o n’s e x p o r t a b l e c o m m o d i t y i si n t e n s i v e i n i t s p r o d u c t i o n,t h e o u t p u t o f t h e c o m m o d i t y w i l l a l s o i n c r e a s e,a n d t h ev o l u m e o f t r a d e w i l l e x p a n s e.幻灯片13●I f t h e c o n s u m p t i o n o f a n a t i o n’s i m p o r t a b l e c o m m o d i t y i n c r e a s e s p r o p o r t i o n a t e l ym o r e t h a n t h e n a t i o n’s E x p o r t a b l e c o m m o d i t y,t h e v o l u m e o f t r a d e a l s o e x p a n s e s.●I f b o t h p r o d u c t i o n a n d c o n s u m p t i o n a r e p r o-t r a d e,t h e v o l u m e o f t r a d e d e f i n i t e l yi n c r e a s e.●I f a n a t i o n’s i n c r e a s e i n l a b o r f o r c e a n d p o p u l a t i o n i s g r e a t e r t h a n i n i t s t o t a l v o l u m eo f c o n s u m p t i o n,t h e n,t h e n a t i o n’s w e l f a r e d e c l i n e s.幻灯片14T e c h n i c a l p r o g r e s s,t r a d e,a n d w e l f a r e●T h e n e u t r a l e x p a n s i o n o f p r o d u c t i o n a n d c o n s u m p t i o n l e a d s t o t h e s a m e r a t e o fe x p a n s i o n of t r a d e.N e u t r a l p r o d u c t i o n a n d p r o-t r a d e c o n s u m p t i o n c a u s eg r e a t e rp r o p o r t i o n a t e e x p a n s i o n o f t h e v o l u m e o f t r a d e t h a n o f p r o d u c t i o n.N e u t r a lp r o d u c t i o n a n d a n t i-t r a d e c o n s u m p t i o n r e s u l t i n a l e s s p r o p o r t i o n a t e e x p a n s i o n o f t h e v o l u m e o f t r a d e t h a n o f p r o d u c t i o n.幻灯片15●H o w e v e r,n o m a t t e r w h a t h a p p e n s t o t h e v o l u m e o f t r a d e,n e u t r a l t e c h n i c a l p r o g r e s sw i l l i n c r e a s e t h e w e l f a r e o f t h e n a t i o n,w i t h c o n s t a n t t e r m s o f t r a d e.●N e u t r a l t e c h n i c a l p r o g r e s s i n t h e p r o d u c t i o n o f t h e e x p o r t a b l e c o m m o d i t y o n l y i sp r o-t r a d e.O n t h e o t h e r h a n d,n e u t r a l t e c h n i c a l p r o g r e s s o n l y i n t h e p r o d u c t i o n o f t h ei m p o r t a b l e c o m m o d i t y i s a n t i-t r a d e.幻灯片16●G i v e n t h e t e r m s o f t r a d e,t a s t e s,a n d p o p u l a t i o n u n c h a n g e d,t h e v o l u m e o f t r a d et e n d t o d e c l i n e,b u t n a t i o n a l w e l f a r e i n c r e a s e s.●I n s u m m a r y,t e c h n i c a l p r o g r e s s,d e p e n d i n g o n t h e t y p e,m a y i n c r e a s e o r d e c r e a s et r a d e,b u t i t w i l l a l w a y s i n c r e a s e s o c i a l w e l f a r e i n a s m a l l n a t i o n.幻灯片17G r o w t h a n d t r a d e:t h e l a r g e-c o u n t r y c a s e●T h e l a r g e-c o u n t r y c a s e r e f e r s t o t h a t c a s e i n w h i c h t h e n a t i o n i s s u f f i c i e n t l y l a r g e t oa f f e c t t h e r e l a t i v e c o m m o d i t y p r i c e s a t w h i c h i t t r a d e s.●G r o w t h a n d t h e n a t i o n’s t e r m s o f t r a d e a n d w e l f a r e:幻灯片18●I f e c o n o m i c g r o w t h,r e g a r d l e s s o f i t s s o u r c e o r t y p e,e x p a n d s t h e n a t i o n’s v o l u m eo f t r a d e a t c o n s t a n t p r i c e s,t h e n,t h e n a t i o n’s t e r m s o f t r a d e t e n d t o d e t e r i o r a t e.O n t h e o t h e r h a n d,i f g r o w t h r e d u c e s t h e n a t i o n’s v o l u m e o f t r a d e a t c o n s t a n t p r i c e s, t h e n a t i o n’s t e r m s o f t r a d e t e n d t o i m p r o v e.T h i s i s r e f e r r e d t o a s t h e t e r m s o f t r a d ee f f e c t o f g r o w t h.幻灯片19●T h e e f f e c t o f g r o w t h o n t h e n a t i o n’s w e l f a r e d e p e n d s o n t h e n e t r e s u l t o f t h e t e r m so f t r a d e e f f e c t a n d a w e a l t h e f f e c t.T h e w e a l t h e f f e c t r e f e r s t o t h e c h a n g e i n t h eo u t p u t p e r w o r k e r o r p e r p e r s o n a s a r e s u l t o f g r o w t h.●A p o s i t i v e w e a l t h e f f e c t,b y i t s e l f,t e n d s t o i n c r e a s e t h e n a t i o n’s w e l f a r e.O t h e r w i s e,t h e n a t i o n’s w e l f a r e t e n d s t o d e c l i n e o r r e m a i n幻灯片20●u n c h a n g e d.●I f t h e w e a l t h e f f e c t i s p o s i t i v e a n d t h e n a t i o n’s t e r m s o f t r a d e i m p r o v e a s r e s u l t o fg r o w t h a n d t r a d e,t h e n a t i o n’s w e l f a r e w i l l d e f i n i t e l y i n c r e a s e.I f t h e y a r e b o t hu n f a v o r a b l e,t h e n a t i o n’s w e l f a r e w i l l d e f i n i t e l y d e c l i n e.I f t h e w e a l t h e f f e c t a n d t h e t e r m s o f t r a d e e f f e c t m o v e i n o p p o s i t e d i r e c t i o n,t h e n a t i o n’s幻灯片21●w e l f a r e m a y d e t e r i o r a t e,i m p r o v e,o r r e m a i n u n c h a n g e d d e p e n d i n g o n t h e r e l a t i v es t r e n g t h o f t h e s e t w o o p p o s i n g f o r c e s.●I m m i z e r i z i n g g r o w t h:E c o n o m i c g r o w t h t h a t r e s u l t s i n a r e d u c t i o n i n n a t i o n a le c o n o m i c w e lf a r e e v e n i f t h e w e a l t h e f f e c t t e n d s t o i n c r e a s e t h e n a t i o n’s w e l f a r e,t h et e r m s o f t r a d e幻灯片22●m a y d e t e r i o r a t e s o m u c h a s t o a n e t d e c l i n e i n t h e n a t i o n’s w e l f a r e.T h i s c a s e w a st e r m e d a s i m m i s e r i z i n g g r o w t h.●I m m i s e r i z i n g g r o w t h m o r e l i k e l y t o o c c u r i n d e v e l o p i n g t h a n i n d e v e l o p e d n a t i o n s.●E v e n t h o u g h t h e t e r m s o f t r a d e o f d e v e l o p i n g n a t i o n s h a v e d e t e r i o r a t e d s o m e w h a t,t h e i r r e a l p e r c a p i t a i n c o m e s a n d w e l f a r e h a v e g e n e r a l l y i n c r e a s e d.幻灯片23案例:欧佩克——防止贫困化的增长有效武器●欧佩克（O P E C），又称石油输出国组织，是一个永久性的政府间组织，也是世界最著名的石油垄断组织。

Lecture 10

can't can't
Exercise There
Key:EDAFB
Key:BC
Key:BCCCD
Homework
1、完成配套习题Lecture10部分 2、复习本课所有学习内容。
Thanks
➢ Sure,I'd love to.
常见表示接受邀请的回答：➢ Yes,I'd love to.
➢ Sure, I think he/she /they will.
➢ Sorry.I'm not available.
常见表示拒绝邀请的回答： ➢ I'm sorry. I'm busy this week. I have many meetings.
Warm-up Have you seen the movie Mission Impossible-Fallout?
New words and expressions
依次读出下列单词
exam;prepare;until;hang;catch
Section A 2d
Role-play Reading
Break Time
Section A 2d
大声读出下列句子
I remember we went bike riding together last fall when he visited you.
Exercise
repare
aught
hung
examined
hang
out prepare for exams not until
4.hang hang out hang up hang on
v.悬挂；垂下闲逛；常去某处挂断电话紧紧抓住

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Translation exercises
《黄帝内经》认为气是生命赖以存在和发展的动力，所谓“正气存内，邪不可干”，“邪之所凑，其气必虚”。关于导引的方法，提到的就有14处之多，而且还总结出了用导引治疗10多种疾病的经验。健身气功锻炼正是通过形体动作、呼吸吐纳和心理调节来引导人体内气运行，从而达到强身健体目的的。《黄帝内经》是中医学四大经典著作之一，是第一部冠以中华民族先祖“黄帝”之名的巨著，是中医现存成书最早的一部医学典籍。是研究人的生理学、病理学、诊断学、治疗原则和药物学的医学巨著，建立了中医学上的“阴阳五行学说”、“经络学说”、“养生学”、 “运气学”等学说。其医学理论反映了中国古代朴素唯物主义辨证思想。
Lesson 10 Wushu and Qigong
A. Wushu
1. What is Wushu? Why is it appealing to many people? 2. How did Wushu come into being? 3. Why has Wushu developed into a great variety of schools and styles? How many schools do you know? 4. Give examples to show the characteristics of various schools of Wushu. 5. What are the 4 main types of Wushu? What are the 3 categories of weapons used in Wushu? 6. Is Wushu limited to the external movement? What is emphasized? What may be an important purpose of Wushu exercises? 7. What’s martial ethics? 8. What’s martial artistry? How did it come into being?
葛洪（284～364或343），道教学者、著名炼丹家、医药学家。字稚川，汉族，江苏句容县人。曾受封为关内侯，后隐居罗浮山炼丹。著有《神仙传》、《抱朴子》等。
Bruce Lee (27 Nov. 1940 – 20 July 1973) , a Chinese American, martial arts instructor, Hollywood actor and film director.
The Yellow Emperor’s Classic of Internal Medicine
Illustrations in The Yellow Emperor’s Classic of Internal Medicine
Huo Yuanjia, a patriotic master of Kungfu
Topics for discussion
1. What martial arts have been listed as sports events of the Asian Games? 2. Why is Taiji Boxing (Shadow Boxing) popular with people of all ages? 3. What Chinese Kungfu movies do you know are/were popular with westerners? Why?
Translation exercises
李小龙(1940 11.27 - 1973 07.20)，一代武术宗师、功夫电影的开创者、武道哲学的创立者和截拳道的创始人。他是将中国功夫传播到全世界第一人，打入好莱坞的首位华人。他革命性地推动了世界武术和功夫电影的发展，创造了世界影迷最多的武术家的世界纪录，全球共拥有本人称他为武之圣者。李小龙的一生是短暂的，但他却创造和打破了多项世界之最、中国之最。他却如同一颗耀眼的彗星划过国际武坛的上空，对现代搏击技击术和电影表演艺术的发展作出了巨大的贡献。他主演的功夫片风行海外，在不少外国人心目中，他的功夫就是中国武术。他还开办“振藩国术馆”，自创截拳道。他用32岁的生命和六部电影缔造了不朽的东方传奇。
A Hollywood star
B. Qigong
1. What else was Qigong called at ancient times? Is it a legacy of Chinese martial arts or Chinese medicine? 2. What does Qi in Qigong mean? What’s internal Qi? 3. What do the channels and collaterals link? What are the main trunks? How are they connected with Qigong? 4. What is Daoyin? How to practice Qigong? 5. What does The Yellow Emperor’s Classic of Internal Medicine say about Qigong? 6. What kinds of Qigong do you know?
Translation exercises
霍元甲（1868－1910）是著名爱国武术家，精武体育会创始人。他有两次打擂的经历。1901年，一俄国人来津，自称“世界第一大力士”，打遍中国无敌手。霍元甲极为气愤，提出要与之决一雌雄。迫于霍元甲的气势，“俄国大力士”灰溜溜地逃离天津。1909年，英国大力士奥比音在上海辱中国人为“东亚病夫”，霍应邀赴上海比武。慑于霍的威名，奥比音未敢交手即消失。1910年，在上海创办精武体育会。孙中山写下“尚武精神”四个大字惠赠。日本柔道会得知霍勇挫俄、英大力士，很不服气, 从国内选派十几名高手，由柔道会长率领来华，请霍来馆比武。失败后，日本人举行宴会招待霍，听说霍咳嗽并在比武中有外伤，就介绍秋野医生为霍治病。霍服药后病情逐渐恶化，后因中毒太深，1910年9月14日长逝。徒弟们拿霍的药去化验，得知是慢性烂肺药，明白是日本人暗下毒手。

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