Interactive Proofs, Proof Strategy
小学英语语法-名词单复数变化规则记忆顺口溜
小学英语语法：名词一．名词的定义：1.名词分为专有名词和普通名词。
专有名词是表示具体的人、事物、地点、或机构等的专有名称。
如：GuangZhou，Mike，UNESCO等。
专有名词一般情况下第一个字母要大写，前面不加定冠词the。
普通名词是表示一类人、事物或抽象概念的名称。
如：police，eggs，rice等。
普通名词又可分为个体名词、集体名词、物质名词和抽象名词。
个体名词：表示某一类人或某一类东西中的个体。
如：monkey，panda，ruler，boy等。
集体名词：表示一群人或一些事物的总称。
如：family，police，class，people等。
物质名词：表示无法分为个体或不具备确定形状和大小的实物。
如：Water，air，milk等。
抽象名词：表示抽象概念词。
如：hope，love，spirit。
二．可数名词和不可数名词：1.定义：一般来说，个体名词和集体名词多为可数名词，物质名词和抽象名词多为不可数名词。
2.可数名词的数：可数名词有单数、复数之分，表示“一个”时用单数，a bird，a teacher，an apple，表示“多个”时用复数。
two birds，five teachers，eight apples3.可数名词复数的变化规则口诀：名词单数变复数，直接加-s 占多数；s, x, z, ch, sh 来结尾，直接加上-es；词尾是 f 或fe，加-s 之前先变ve；辅母+ y 在词尾，把y 变i 再加-es；词尾字母若是o，常用四个已足够，要加-es 请记好，黑人英雄吃土豆西红柿. A、一般情况下，在单数名词的后面加-s构成。
如：game-games，boy-boys等。
B、以s、x、sh、ch结尾的单数名词变复数，在词尾加-es构成。
如：box-boxes，bus-buses，peach-peaches，dish-dishes。
C、以o结尾，表示有生命的事物的单数名词变复数加-es；口诀：黑人和英雄吃土豆和西红柿+esnegroes and heroes eat tomatoes and potatoes表示没有生命的事物的单数名词变复数加-s。
多邻国词汇(高阶)
A1.and 和，及2.area 区域(areas 复数)3.amphibians 两栖动物4.all 所有的5.aircraft 飞机6.automobile 发动机（汽车）7.agreement 一致8.appear 出现9.actual 真实的10.aliens 外侨11.argued 争论(argue 的过去分词）12.amino 氨基13.availability 可获得性14.app 应用程序(apps 复数)15.absolute 绝对的16.always 常常17.along 一起(along with...）18.analysis 分析19.adopted 领养的20.anyone 任何人21.abolitionist 废奴主义者22.approximately 大约23.American 美国的24.asteroid 小行星(asteroids 复数）25.addition 添加(in addition 此外)26.almost 几乎27.achievement 成就28.against 违背29.attract 吸引30.acid 酸的31.anonymity 匿名32.accountable 负有责任的33.access 访问(accessed 过去式)34.analyze 分析analyzed 过去式/过去分词) 35.apartment公寓36.adverse有害的（have adverseeffect on对…产生有害的影响）37.addition 增加(in addition to 除…之外）38.allege 声称（过去式alleged)39.agent 特工40.aquatic 水生的(aquatic environment 水生环境）41.artificial 人工的42.allow 允许（allowing 现在分词）43.attach 附加（过去分词attached)44.although 即使45.allergy 过敏B1.be 是/用于被动语态2.boundary 边界(boundaries 复数）3.by 通过4.before 在… 之前5.badger 獾6.bank 银行7.believed 相信(believe 的过去分词）8.beads 小珠子(bead 的复数）9.build 建立(built 过去式）10.back 后退11.beside 在…旁边12.behavior 行为13.base 基地14.basic 基础的15.blend 混合16.buzz 发出嗡嗡声(buzzed 过去式/过去分词)17.bird 鸟18.beneath 在…下面19.blow 重击20.block 街区21.belong to 属于22.battered 饱受折磨的23.build upon 建立于…C1.cellular 细胞的2.classroom 教室3.camera 相机(cameras 复数)4.capability 能力5.cheat 作弊6.calculate 计算(calculations 名词复数）7.check 检查(checks 单数第三人称)"8.call 称为9.control 控制controlled 过去分词10.consist (of) 由…构成11.concerned 关注的关心的12.certain 必然的13.core 核心14.construction 建造puter 电脑16.carnivores 食肉动物17.coyote 丛林狼18.Chinese 中国的19.camp 露营(camps 复数)20.cycle 自行车bustion 燃烧22.captive 俘虏23.conjure 想象(conjure up 想象出）(conjures 单数第三人称）puterized 电脑化的25.changes 变化(change 的复数)(changed 过去式/过去分词)26.content 内容27.church 基督教堂(churches 复数)28.country 国家29.case 案件(cases 复数）30.copyright 版权pany 公司32.cluttered 杂乱的33.counter 柜台(counters 复数) 34.cover 覆盖(covered 过去分词，可做形容词）35.centric 重要的36.contend 声称(contended 过去分词）37.campaign 竞选38.chance 机会et 彗星(comets 复数）40.crisis 危机41.cognitive 认知的42.calligraphy 书法43.cursive 草书44.cloud 云ponent 组件(components 复数）46.calve 小腿47.crowd 人群48.color 颜色(colored 有颜色的)49.curls卷发50.convert 转换(converting 现在分词）51.code 代码52.camouflage 伪装bined 综合的（combinedeffects 综合效应）54.contention 观点55.curve 曲线56.conspiracy 阴谋57.classical 古典的（classical school古典学派）58.credential证书授权59.check 检查60.card 身份证；卡61.critical 批判的62.cinematography电影摄影学63.call 调用（program calls 程序调用）64.create 创造65.consumer 消费者（consumerproduct 消费品）pensate 补偿（compensate for)67.correctly 准确地puter电脑69.cover 涉及70.celiac 腹部的（celiac disease)D1.driven 驱使2.durable 耐用品3.delay 推迟(delayed 过去分词)4.diesel 柴油5.determine 决定6.decimate 大大削弱7.design 设计(designed 过去式/过去分词)8.dominion 主权，统治权9.decrepit 破旧的10.dirty 脏的11.dust 灰尘12.department 部门(departments 复数)13.dwarf 矮种的14.death 死亡15.density 密度16.differently 不同地17.demining 排雷18.detect 查明19.dose 剂量(dosing 定量)20.(doses 单数第三人称)"21.data 数据22.defense 防御23.dealer 发牌者24.disagree 不同意（disagree with）25.definitive 明确的26.determinism 决定论（culturaldeterminism文化决定论）27.destroy 破坏28.dwarfism 侏儒症29.dysplasia发育不良30.data 数据31.database数据库32.duplicate 复制33.devolve 移交34.disorder 疾病E1.equilateral 等边的(equilateral triangle 等边三角形）2.engine 发动机3.especially 特别是4.expressly 明确地5.environmental 环境的6.external 外部的7.example 例子8.effect 影响(negative effect 副作用）9.enforced 强制执行cation 教育11.enter 进入12.ever 曾经13.exception 例外14.enzyme 酶(enzymes 复数)15.electromagnet 电磁铁16.equal 相同的17.examine 检测（过去式examined)18.exist 存在19.exactly 确切地20.environment 环境21.entire 整个的22.enormous 巨大的23.engineering工程（softwareengineering 软件工程）24.equipment 设备F1.factor 因素2.feature 特点(features 复数）(feature …of 以…为特色的）3.fast 快的(faster 比较级）4.focusing 聚焦(focus 的现在分词)(focus on 关注于)"5.functionality 功能性6.ferret 雪貂7.formal 正规的8.fall 秋天9.federal 联邦的10.function 功能11.flipchart 挂图12.following 关注(follow 的现在分词）13.farther 较远的14.floor 地板15.feudalism 封建制度16.field 领域17.fearing 害怕18.final 最终的19.front 前面的20.flexible 灵活的21.floor 底层（sea floor海底）22.fin 鳍(fins 复数）23.frog 青蛙24.funding 资金（funding source资金源）25.fitted 合身的26.flare out 展开27.focus on集中于28.facilitate 促进G1.greatest 最伟大的ernment 政府3.growing 增长的4.general 一般的5.group 组6.grape 葡萄7.graph 图表8.figure 图形9.giant 巨人H1.have 拥有2.hole 洞3.home 家4.herder 牧人(herders 复数）5.headed 率领6.hour 小时7.hot 热(hotter 比较级）"8.hind 后部的9.hydrogen 氢气10.happen 发生11.hit 打，击12.hotel 旅馆13.hunchback 驼背14.hard work 努力工作15.historic 具有历史意义的16.housing 建筑I1.internet 互联网2.is 是3.includes 包含4.implementation 实行，实施5.intense 强烈的6.industry 行业7.infringement 侵犯8.independent 独立的9.interest 兴趣(interested 感兴趣的)(be interested in …)10.internal 内部的11.ideology 意识形态12.inkling 暗示13.impact 影响(impact on …)14.inherent 固然的15.infidelity 不忠16.including 包括17.ionized 电离的18.idea 想法19.identify 识别(identified 过去式/过去分词）20.inning 局次21.invent 捏造22.individual 个人23.impact 影响24.inspire 启发25.interface 交互；接口26.implementation执行J1.Japanese 日本的2.join 加入3.journal 期刊4.jelly 果冻K1.known 已知的2.knife 刀L1.live 生活2.like 例如3.line 行4.lead 通往leader 领导5.leave 离开(leaves 单数第三人称）6.linear 线性的7.letter字母letters 复数8.license 许可证9.leathery 皮质的10.litter 垃圾11.literacy 文化12.learn 学习(learning 现在分词，可用作名词）13.lowering 降下14.level 水平; 齐平的(levels 复数)15.loyalty 忠诚tter 后者的nd 土地18.low 低的（low taxes 低税）19.limiting 限制的20.limb 肢体（limbs 肢体）21.line 线条（line segment 线段）22.lost 消失的23.local 当地的M1.messaging 信息收发2.many 许多(more 比较级)3.memory 存储器4.make 使(made 过去式)5.modern 现代的6.manual 手动的7.mink 貂皮8.mesh 网状结构的9.menu 目录10.model 模型，做… 的模型11.manner 方式12.morning 早上13.mixture 混合14.mosque 清真寺(mosques 复数)15.Mormons 摩门教徒16.mess 杂乱17.major 主要的18.molecular 分子的19.medium 媒介20.mean 意味着21.mate 交配mating (现在分词/名词)22.measure测量23.muscular 肌肉的24.majority 大部分25.melt 融化26.material 材料27.methodology 方法论28.manner 态度（bedside manner医生对病人的态度）29.marcher 游行者N1.not 不，表否定2.negotiate 洽谈3.number 数量 名字5.nearly 将近6.nominee 候选人7.nausea 呕吐感8.north 北部的 网民10.natural 自然的11.nominally 名义上地O1.other 其他2.of 属于,…的3.occupy 占据4.object 目标; 物品（objects 复数）5.order 秩序(ordered 有秩序的）anization 组织(organizations 复数)7.orally 口服8.only 仅，只有9.other 其他的10.original 原本的(originally 原本地)11.output 输出(outputs 单数第三人称)12.oak 橡树13.old 古老的14.outright 彻底的15.override 覆盖anize 组织17.office 部门P1.particular 格外的(in particular 尤其是)2.program 程序programmer 程序员3.proof 证据(proofs 单数第三人称)4.percent 百分比5.psychology 心理学6.petrochemicals 石油化工产品7.practical 实际的, 行动的8.provide 提供(provides 单数第三人称）9.plan 计划10.press 按，压11.pop 突然出现(pop up 弹出的）12.party 政党13.power 权利14.property 性能，财产(properties 复数）15.picture 图片16.professional 职业的17.printed 印刷的18.protect 保护19.publish 发布20.problem 问题，困难(problems 复数）21.parliament 议会22.present 展现23.passageway 通道走廊24.probably 大概25.participatory 参与式的26.previous 在…之前的27.poll 投票(polls 复数)28.protein 蛋白质(proteins 复数）29.place 放置(placed 过去分词）30.prefer 偏爱31.plasma 等离子32.prevent 阻止(prevented 过去式/过去分词）33.proclaimed 公告34.particular 特别的35.push 逼迫36.packing 包装37.perform 执行38.precaution 预防措施39.platform 平台40.profiler 分析器41.predator 捕食者(predators 复数）42.plant 种植(planted 过去式）43.parasite 寄生虫44.preparation 准备45.proxy 代理46.process 过程47.propaganda宣传48.plot 故事49.purpose 目的50.pressure 压力51.paring 削皮（paring knife 削皮刀）52.program 程序53.path 小路R1.require 要求(requires 单数第三人称)2.representative 有代表性的3.religion 宗教，信仰4.response 反应5.reasoning 推理，疑问6.reptile 爬行动物7.range 范围8.red 红色9.river 河流10.reluctantly 不情愿的11.rights 权利12.refocus 重新关注13.replaced 替代(replace 的过去)14.religious 信仰的15.reflect 反映(reflects 单数第三人称)16.relatively 相对地17.return 还回18.result 导致(result in）19.realism 现实主义20.regardless 无论21.restless 好动的22.rebellious 叛逆的23.repel 排斥24.reprisal 报复25.rule 规则(rules 复数)26.reproduction 繁殖27.root 根茎28.room 房间29.resort 旅游胜地30.rich 富裕的（richest 最高级）31.remove 移除32.ring 环33.randomize 随机化34.repeat 重复35.replace 替换（be replaced with）36.recover 恢复37.right正好，就在38.redundant 冗余的39.rainforest 雨林40.religion宗教41.refuse 拒绝42.reader 读卡器43.resemble 像44.route道路S1.so 因此2.symbol 符号3.second 第二4.species 物种5.steel 钢铁6.skunk 臭鼬7.summer 夏天8.seek 寻找(sought 过去式/过去分词）9.split 分裂10.speed 速度11.sentence 句子(sentences 复数12.sequence 顺序13.synagogue 犹太教会堂(synagogues 复数）14.significance 重要性15.star 恒星(stars 复数)16.sloping 倾斜的17.service 服务，检修(services 第三人称单数）18.start 开始19.spot 斑点(spots 复数）20.serrated 锯齿状的21.still 静止22.supposedly 据说23.strategy 战略(strategies 复数）24.sinusitis 鼻窦炎25.system 系26.sober 清醒的27.soluble 可溶的28.script 手稿29.style 风格30.size 大小31.sensor传感器(sensors 复数)32.south 南部的33.sudden 突然的34.simple 简单的35.sphere 球体36.supposed 假定的37.swallow 吞咽38.serial 系列(serials 复数)39.store 存储(stored 过去式/过去分词）"40.survival 存活41.stretch 延伸(stretched 过去式/过去分词)42.segment 部分43.statue 雕塑（bronze statues 青铜雕塑）44.soil 泥土45.scientist科学家46.soften 缓和47.successor 继任者48.shelf 架子（shelves 复数）49.several 若干个50.sacrifice 牺牲（make sacrifice)51.suffer from 遭受52.speedy 快速53.shock 使…震惊54.support 支持55.stable 稳定的56.shift 改变57.structural 结构的58.symptom症状59.sensitivity 过敏T1.the 这/那；用在最高级之前2.tendency 趋势3.theoretical 理论的4.troop 军队(troops 复数) 5.trade 交易6.term 关系，术语(in terms of 依据…)7.treat 零食,对待(treatment 名词）(treats 复数）"8.tunnel 隧道9.thousand 千(thousands 复数)(thousands of 数千的）10.tissue 组织11.target 目标12.toxic 有毒的(toxicity 毒性)(toxicities 复数)13.tape 磁带14.tropical 热带的15.temperate 温和的16.tag 标签17.transfer 转移U1.understanding 理解能力e 使用3.unionist 工会会员4.undercover 秘密的5.unstable 不稳定的6.upsets 挫折ual 通常的8.uphill 上坡的9.underside 底部10.utensil 厨房用具V1.view 视图2.variation 变化，差异3.value 价值4.video 视频(videos 复数）5.vehemently 激烈地6.vegetation 植物7.vulnerable 脆弱的8.vernal 春天的9.vast 巨大的10.very 恰好的11.vehicle交通工具W1.work 工作；作品(work out 算出)2.with 和…，表伴随3.weasel 黄鼠狼4.west 西部5.whiteboard 白板6.wing 翅膀(wings 复数)7.warn 警告8.window 窗户(windows 复数）9.word 字10.width 宽度11.wheat 小麦Y1.young 年轻的。
薄冰英语语法知识点
薄冰英语语法知识点薄冰英语语法：句子成分句子成分共有六种：1)主语(subject)是一句的主体，是全句述说的对象，常用名词或相当于名词的词担任，一般置于句首。
如：(2)The sun is shining in our faces. 阳光正照射在我们脸上。
shine [ʃain] vi. 发出光;反射光，闪耀;出类拔萃，表现突出;露出;照耀;显露;出众vt. 照射，擦亮;把…的光投向;(口)通过擦拭使…变得有光泽或光n. 光亮，光泽;好天气;擦亮;晴天;擦皮鞋;鬼把戏或诡计shining ['ʃainiŋ] v. 闪烁(shine的ing形式);发亮adj. 光亮的;(3)I respect his privacy. 我尊重他的隐私权。
respect [ri'spekt] n. 尊敬，尊重;方面;敬意vt. 尊敬，尊重;遵守 privacy ['privəsi; 'prai-] n. 隐私;秘密;隐居;隐居处2)谓语或谓语动词(predicate or predicate verb)是说明主语的动作或状态的，常用动词担任，置于主语之后。
如：当马丁离开水果店时，黄昏开始降临。
twilight ['twailait] n. 黎明，黄昏;薄暮;衰退期;朦胧状态 adj. 昏暗的，微明的;暮年的(5)The play began [bi'ɡæn] at eight, so they must dine at seven. 戏8点钟开演，所以他们必须7点钟吃饭。
begin at 几点开始;从……开始dine [dain] vi. 进餐，用餐vt. 宴请“谓语或谓语动词”专指动词部分(包括动词短语)。
它与“谓语部分”不同，二者不可混淆。
3)宾语(object)是表示及物动词的动作对象和介词所联系的对象的，常由名词或相当于名词的词担任。
置于及物动词或介词之后。
[全]新概念英语 第三册 第30课第⑦句
![[全]新概念英语 第三册 第30课第⑦句](https://img.taocdn.com/s3/m/6d675eee77232f60ddcca1cb.png)
新概念英语第三册第30课第⑦句第三段词汇句型语法解析及翻译第⑦句⑦The only other people who knew the secret were Joe and Bert. 除此之外，只有乔与鲍知道这个秘密。
句子结构分析:注意，句子only的意思不是说“只有一个人”，而表示“仅仅”。
Who引导定语从句修饰the only other people，此处的先行词people由关系代词WHO 来修饰的一般不可以用that来代替。
这个语法知识点是词句，乃至全课文里最后语法点，请研读感悟以下分析:关于定语从句中先行词由only修饰其关系代词选用的几种情况:1.如果the only修饰的先行词是物，事情，切记定语从句的关系代词必须是that 来引导，而不能用which。
2.the only + 先行词其后所接的定语从句用什么关系代词来引导定语从句，通常有一个误区。
有的老师简单的告诉学生，只要先行词是被the only修饰的，一概要用that关系代词，这种说法是不完全对的。
3.正确的说法是，当受定语从句修饰的先行词指人，且受到the only修饰时，定语从句用who关系代词来引导是比较正式的。
当受定语从句修饰的先行词指人，且受到the only修饰时，定语从句通常必须用关系代词who来引导的几种情况：He’s the only one who I know. 他是唯一我认识的人。
Ha Wen is the only person who can do this kind of survey. 哈文是唯一会做这种问卷调查的人。
I’m the only one who’s gone to a public high school. 我是惟一上过公立中学的人。
（《薄冰高级英语语法》）Sally is the only person in our class who dares (to) answer Miss Thompson back. 在我们班上，只有萨利一个人敢和汤普森小姐顶嘴。
On the Composition of Zero-Knowledge Proof Systems
On the Composition of Zero-Knowledge Proof SystemsOded Goldreich Hugo Krawczyk*Dept.of Computer Science IBM T.J.Watson Research CenterTechnion,Haifa Yorktown Heights,NY10598Israel U.S.A.ABSTRACT:The wide applicability of zero-knowledge interactive proofs comes from the pos-sibility of using these proofs as subroutines in cryptographic protocols.A basic question con-cerning this use is whether the(sequential and/or parallel)composition of zero-knowledge proto-cols is zero-knowledge too.We demonstrate the limitations of the composition of zero-knowledge protocols by proving that the original definition of zero-knowledge is not closed under sequential composition;and that even the strong formulations of zero-knowledge(e.g.black-box simulation)are not closed under parallel execution.We present lower bounds on the round complexity of zero-knowledge proofs,with significant implications to the parallelization of zero-knowledge protocols.We prove that3-round interac-tive proofs and constant-round Arthur-Merlin proofs that are black-box simulation zero-knowledge exist only for languages in BPP.In particular,it follows that the"parallel versions"of thefirst interactive proofs systems presented for quadratic residuosity,graph isomorphism and any language in NP,are not black-box simulation zero-knowledge,unless the corresponding languages are in BPP.Whether these parallel versions constitute zero-knowledge proofs was an intriguing open question arising from the early works on zero-knowledge.Other consequences are a proof of optimality for the round complexity of various known zero-knowledge protocols, and the necessity of using secret coins in the design of"parallelizable"constant-round zero-knowledge proofs.Research was partially supported by the Fund for Basic Research Administered by the Israeli Academy of Sciences and Humanities.Preliminary version appeared in the Proc.of the17th ICALP,Lecture Notes in Computer Science,Vol.443,Springer Verlag,pp.268-282,1990.*Work done while the author was with the Department of Computer Science,Technion,Haifa, Israel.-2-1.INTRODUCTIONIn this paper we investigate the problem of composing zero-knowledge proof sys-tems.Zero-knowledge proof systems,introduced in the seminal paper of Goldwasser, Micali and Rackoff[GMR1],are efficient interactive proofs which have the remarkable property of yielding nothing but the validity of the ly,whatever can be efficiently computed after interacting with a zero-knowledge prover,can be efficiently computed on input a valid assertion.Thus,a zero-knowledge proof is computationally equivalent to an answer of a trusted oracle.A basic question regarding zero-knowledge interactive proofs is whether their com-position remains zero-knowledge.This question is not only of theoretical importance,but is also crucial to the utilization of zero-knowledge proof systems as subprotocols inside cryptographic protocols.Of particular interest are sequential and parallel composition. Candidate"theorems"(whose correctness we investigate)are:Sequential Composition:LetΠ1andΠ2be zero-knowledge proof systems for languages L1and L2respectively.Then,on input x1 x2,executingfirstΠ1on x1and afterwards executingΠ2on x2,constitutes a zero-knowledge interactive proof system for L1 L2. Parallel Composition:LetΠ1andΠ2be as above.Then,on input x1 x2,executing con-currentlyΠ1on input x1andΠ2on x2,constitutes a zero-knowledge interactive proof system for L1 L2.(Concurrent execution means that the i-th message of the composed protocol consists of the concatenation of the i-th messages inΠ1andΠ2,respectively).Sequential CompositionSoon after the publication of the[GMR1]paper,several researchers noticed that the formulation of zero-knowledge proposed therein(hereafter referred as the original for-mulation)is probably not closed under sequential composition.In particular,Feige and Shamir[Fei]proposed a protocol conjectured to be a counterexample to the Sequential Composition"Theorem".In this paper we use the ideas of[Fei]and new results on pseu-dorandom distributions[GK],to prove that indeed the original formulation of zero-knowledge is not closed under sequential composition.Our proof is independent of any intractability assumption.It applies to the notion of computational zero-knowledge(see Section2),and uses computationally unbounded provers.(So far no proof exists for the same result with provers limited to polynomial-time,or for statistical or perfect zero-knowledge.)The reader should be aware that the Sequential Composition Theorem was proven (by Goldreich and Oren[GO,Ore])for a stronger("non-uniform")formulation of zero-knowledge suggested by several authors(cf.[Fei,GMR2,GO,Ore,TW]).The Sequen-tial Composition Theorem is crucial to the utilization of zero-knowledge interactive proofs in cryptographic applications and in particular to the construction of cryptographic protocols for playing any computable game[Yao,GMW2].Parallel CompositionParallel composition of interactive proofs is widely used as a means of decreasing the error probability of proof systems,while maintaining the number of rounds.Of course one would be interested in applying these advantages of parallelism to zero-knowledge protocols as well.Parallelism is also used in multi-party protocols in which parties wish to prove(the same and/or different)statements to various parties con-currently.Unfortunately,we show in this paper a counterexample to the Parallel Compo-sition"Theorem".Namely,we introduce a pair of protocols which are(computational) zero-knowledge with respect to the strongest known definitions(including the non-uniform formulation discussed above and the"black-box simulation"formulation dis-cussed bellow)yet their parallel composition is not zero-knowledge(not even in the "weak"sense of the original[GMR1]formulation).Also in this case,our proof does not rely on any unproven hypotheses;on the other hand it uses in an essential way the unbounded computational power of the prover and the computational notion of zero-knowledge.Based on intractability assumptions,Feige and Shamir[FS2]show a perfect zero-knowledge protocol with a polynomial-time prover which fails parallel composition. Our results below on3-round zero-knowledge proofs imply a similar result but our case requires a super-logarithmic number of repetitions while in[FS2]two repetitions suffice.By the above result we have ruled out the possibility of proving that particular interactive proofs are zero-knowledge by merely arguing that they are the result of paral-lel composition of various zero-knowledge protocols.But this does not resolve the ques-tion whether concrete cases of composed interactive proofs are zero-knowledge.In par-ticular,since thefirst presentation of the results in[GMR1]and[GMW1]it was repeat-edly asked whether the"parallel versions"of the interactive proofs presented for Qua-dratic Residuosity,Graph Isomorphism and for any language in NP are zero-knowledge.Round complexity of zero-knowledge proofsIn this paper we prove a general result concerning the round complexity of zero-knowledge interactive proofs which,in particular,resolves the question of parallelization of the above mentioned protocols.This general result states that only BPP languages have3-round interactive proofs which are black-box simulation zero-knowledge.1Since1This result applies to interactive proofs in which the prover can convince the verifier ofthe parallel versions of the above examples are3-round interactive proofs(with negligi-ble cheating probability for the prover)it follows that these interactive proofs cannot be proven zero-knowledge using black-box simulation zero-knowledge,unless the corresponding languages are in BPP.This(negative)result is proven for computational zero-knowledge proofs and therefore applies to statistical and perfect zero-knowledge as well.Loosely speaking,we say that an interactive proof for a language L is black-box simulation zero-knowledge if there exists a(probabilistic polynomial-time)universal simulator which using any(even non-uniform)verifier V*as a black box,produces a pro-bability distribution which is polynomially indistinguishable from the distribution of conversations of(the same)V*with the prover,on inputs in L.This definition of zero-knowledge is more restrictive than the original one which allows each verifier V*to have a specific simulator S V*.Nevertheless,all known zero-knowledge protocols are also black-box simulation zero-knowledge.This fact cannot come as a surprise since it is hard to conceive of a way of taking advantage of the full power of the more liberal definition.It is not plausible that our result could be extended to4-round interactive proofs since such proofs are known for languages believed to be outside BPP.The protocols for Quadratic Non-Residuosity[GMR1]and Graph Non-Isomorphism[GMW1]are such examples.In addition,zero-knowledge interactive proofs of5rounds are known for Qua-dratic Residuosity and Graph Isomorphism[BMO1],and assuming the existence of claw-free permutations there exist5-round zero-knowledge interactive proofs for any language in NP[GKa].Moreover,our results extend to zero-knowledge arguments2,for which Feige and Shamir[FS]presented(assuming the existence of one-way functions)a 4-round protocol for any language in NP.Our result implies that the round complexity of this protocol is optimal(as long as BPP≠NP).accepting a false assertion with only negligible probability.The above mentioned languages have3-round zero-knowledge interactive proofs in which the prover has a significant(e.g.constant) probability of cheating.2Interactive arguments(also known as"computationally sound proofs"and"computationally convincing protocols")differ from an interactive proof system in that the soundness condition ofthe system is formulated with respect to probabilistic polynomial-time provers,possibly with auxiliary input(see[BCC]).Namely,efficient provers cannot fool the verifier into accepting aninput not in the language,except with negligible probability.Constant Round Arthur-Merlin ProofsWhen restricting ourselves to Arthur-Merlin interactive proofs,we can extend the above result to any constant number of rounds.We show that only BPP languages have constant-round Arthur-Merlin proofs which are also black-box simulation zero-knowledge.Arthur-Merlin interactive proofs,introduced by Babai[Bab],are interactive proofs in which all the messages sent by the verifier are the outcome of his coin tosses.In other words,the verifier"keeps no secrets from the prover".Our result is a good reason to believe that the only feasible way of constructing constant-round zero-knowledge interactive proofs is to let the verifier use"secret coins".Indeed,the above mentioned constant-round zero-knowledge proofs,as well as the constant round statistical zero-knowledge proofs of[BMO2],use secret coins.Thus,secret coins do help in the zero-knowledge setting.This should be contrasted with the result of Goldwasser and Sipser [GS]which states that Arthur-Merlin interactive proofs are equivalent to general interac-tive proofs(as far as language recognition is concerned).They show that any language having a general interactive proof of k rounds,has also an Arthur-Merlin proof of k ing our result we see that in the zero-knowledge setting such a preservation of rounds(when transforming IP into AM)is not plausible(e.g.,Graph Non-Isomorphism).Our result concerning Arthur-Merlin proofs is tight in the sense that the languages considered above(e.g.Graph Non-Isomorphism,every language in NP)have unbounded (i.e.ω(n)-round,for every unbounded functionω:N→N)Arthur-Merlin proof systems which are black-box simulation zero-knowledge.In particular,we get that bounded round Arthur-Merlin proofs which are black-box zero-knowledge exist only for BPP, while unbounded round proofs of the same type exist for all PSPACE(if one-way func-tions exist[IY,B*,Sha]).That is,while thefinite hierarchy of languages having black-box zero-knowledge Arthur-Merlin proofs collapses to BPP(=AM(0)),the correspond-ing infinite hierarchy contains all of PSPACE.This implies(assuming the existence of one-way functions)a separation between the two hierarchies.Organization:In Section2we outline the definitions of interactive proofs and zero-knowledge,and introduce some terminology and notation used through the paper.Sec-tion3presents the definitions and results concerning pseudorandom distributions that are used for disproving the composition theorems.In Sections4and5we present these dis-proofs for the case of sequential and parallel composition,respectively.Finally,in Sec-tion6we present the lower bounds on the round complexity of black-box simulation zero-knowledge proofs.We stress that this last section is technically independent fromsections3,4and5,and can be read without going through these sections.2.PRELIMINARIESThe notions of interactive proofs and zero-knowledge were introduced by Goldwasser,Micali and Rackoff[GMR1].Here,we give an informal outline of these notions.For formal and complete definitions,as well as the basic results concerning these concepts,the reader is referred to[GMR1,GMW1,GMR2].An interactive proof is a two-party protocol in which a computationally unrestricted prover,P,interacts with a probabilistic polynomial-time verifier,V,by exchanging mes-sages.Both parties share a common input x.At the end of the interaction the verifier computes a predicate depending on this input and the exchanged messages in order to accept or reject the input x.Such a protocol,denoted<P,V>,is called an interactive proof for a language L if the following two conditions hold:Completeness property:For any positive constant c and sufficiently long x∈L, Prob(V accepts x)>1−|x|−c.Soundness property:For any positive constant c and sufficiently long x∈|L, Prob(V accepts x)<|x|−c,no matter how the prover behaves during the protocol.(The above probabilities are taken over the coin tosses of both the prover and verifier).In other words we require that on inputs belonging to L the probability that the prover P "convinces"V to accept the common input is almost1,while on inputs outside L there is no prover that can fool V into accepting,except with negligible probability.Note:Notice that we define an interactive proof to have a negligible probability of error. Some authors define this probability to be just a constant(e.g.1/3).The latter is motivated by the fact that constant error interactive proofs can be converted into negligi-ble error proofs by parallel repetition.However,in the setting of zero-knowledge interac-tive proofs our results show that such parallel repetition may sacrifice the zero-knowledge property.An interactive proof in which the honest verifier chooses all its messages at random(i.e. with uniform probability over the set of all strings of same length as the message)is called an Arthur-Merlin interactive proof[Bab].That is,in an Arthur-Merlin proof sys-tem the only non-trivial computation carried out by the honest verifier is the evaluation of a polynomial-time predicate at the end of the interaction,in order to decide the accep-tance or rejection of the input to the protocol.We say that such a verifier uses public coins.(Notice that there is no"public coin"restriction on the cheating verifiers).We say that an interactive proof has k rounds if there are a total of k messages(alter-nately)exchanged between the prover and verifier during the protocol(i.e.we count mes-sages from both parties).In general,the number k can be a function k(|x|)of the input length.The notation IP(k)stands for the class of languages having k-round interactive proofs,and AM(k)for languages having k-round Arthur-Merlin interactive proofs.An interactive proof is called zero-knowledge if on input x∈L no probabilistic polynomial-time verifier(i.e.one that may arbitrarily deviate from the predetermined program)gains information from the execution of the protocol,except the knowledge that x belongs to L.This means that any polynomial-time computation based on the conversation with the prover can be simulated,without interacting with the real prover, by a probabilistic polynomial-time machine("the simulator")that gets x as its only input. More precisely,let<P,V*>(x)denote the probability distribution generated by the interactive machine(verifier)V*which interacts with the prover P on input x∈L.We say that an interactive proof is zero-knowledge if for all probabilistic polynomial-time machines V*,there exists a probabilistic expected polynomial time algorithm M V*(called the simulator)that on inputs x∈L produces probability distributions M V*(x)that are poly-nomially indistinguishable from the distributions<P,V*>(x).(This notion of zero-knowledge is also called computational zero-knowledge).3The above formalization of the notion of zero-knowledge is taken from the original paper by Goldwasser,Micali and Rackoff[GMR1].Later,stronger formulations of zero-knowledge were introduced in which the simulation requirement is extended to deal with stronger verifiers[Fei,GMR2,GO,Ore,TW].Namely,verifiers with non-uniform properties,e.g.probabilistic polynomial-time verifiers which get an additional auxiliary-input tape,or verifiers modeled by polynomial-size circuits.One further formulation of zero-knowledge is called black-box simulation zero-knowledge[GO,Ore].This formulation differs from the former by requiring the existence of a("universal")simulator that using any(non-uniform)verifier V*as a black-box,succeeds in simulating the<P,V*>interaction.In other words,there exists a probabilistic expected polynomial time oracle machine M such that for any polynomial size verifier V*and for x∈L,the distributions<P,V*>(x)and M V*(x)are polynomially indistinguishable.3Other definitions were proposed in which it is required that the distribution generated by the simulator is identical to the distribution of conversations between the verifier and the prover (perfect zero-knowledge),or at least statistically close(statistical zero-knowledge).See[GMR2]for further details.Remark:A complete formalization of the notion of black-box simulation zero-knowledge requires dealing with the following technical problem.The simulator uses V* as a black-box.This means that the simulator is responsible,during the simulation pro-cess,of feeding into the black-box the external parameters that determine the behavior of V*.These parameters are the string representing the input to the protocol,the contents of the random tape of V*,and the messages of the prover.A problem arises when feeding the random coins used by V*.Although the number of coin tosses used by a particular verifier V*is bounded by a polynomial,there is no single polynomial that bounds this number for all possible verifiers.On the other hand,the simulator M runs(expected) time that is bounded by a specific polynomial.So,how can this simulator manage to feed a verifier requiring more coin tosses than this bound?In[BMO2]this problem is over-come by stating the existence of two random tapes for M.Thefirst is used in the regular way for M’s computations.The second can be entirely fed by M into V*in a single step. That is,M can feed the random coins for the black-box in an"intelligent way"as long as the number of coins does not exceed the time capability of M,beyond this number it can only feed truly random bits.We stress that this formalization is general enough to include all known zero-knowledge proofs.An alternative solution to the above problem is to have,for each polynomial p,a simula-tor M p which simulates all verifiers V*that use at most p(|x|)random coins on any input x.Clearly,the running time of the simulator M p may depend on the polynomial p and then the above difficulty is overcome.This second formulation is weaker than the one proposed in[BMO2],but it suffices for the results proved in our paper and therefore adopted here.(In fact our results of Section6only require the existence of a simulator that simulates deterministic verifiers,i.e.M p with p≡0).Based on the above remark we give our definition of black-box simulation zero-knowledge.Definition:An interactive proof<P,V>is called black-box simulation zero-knowledge if for every polynomial p,there exists a probabilistic expected polynomial time oracle machine M p such that for any polynomial size verifier V*that uses at most p(n)random coins on inputs of length n,and for x∈L,the distributions<P,V*>(x)and M p V*(x)are polynomially indistinguishable.Note:The notion of polynomial indistinguishability in the above definition can be for-malized based on non-uniform polynomial size distinguishers,or uniform polynomial-time distinguishers which have black-box access to the corresponding V*.Our results apply to both formalizations.Terminology:Through this paper we use the term negligible for denoting functions that are(asymptoticly)smaller than one over any polynomial,and the term non-negligible for denoting functions that are greater than one over somefixed polynomial.More precisely, a function f from nonnegative integers to reals is called negligible if for all constants c and sufficiently large n,f(n)<n−c.The function f is called non-negligible if there exists a constant c such that for all(sufficiently large)n,f(n)>n−c.(Observe that non-negligible is not the complement of negligible).Notation:We use the notation e∈R S for"the element e is chosen with uniform probabil-ity from the set S".3.ON EVASIVE AND PSEUDORANDOM SETSIn the demonstration of counterexamples for the"composition theorems"we make use of pseudorandom distributions which have some interesting"evasiveness"properties. These properties and the corresponding proofs are given in[GK]and cited here without proof.Roughly speaking,a distribution on a set of strings of length k is pseudorandom if this distribution cannot be efficiently(i.e.in time polynomial in k)distinguished from the uniform distribution on the set of all strings of length k.In order to formalize this notion one has to define it in asymptotical terms and refer to collections of distributions(called pseudorandom ensembles),rather than single distributions.The notion of a"pseudoran-dom set"is made precise in the following definition.Definition3.1:A set S⊆{0,1}k is called(τ(k),ε(k))-pseudorandom if for any(proba-bilistic)circuit C of sizeτ(k)with k inputs and a single output|p C(S)−p C({0,1}k)|≤ε(k)where p C(S)(resp.p C({0,1}k))denotes the probability that C outputs1when given ele-ments of S(resp.{0,1}k),chosen with uniform probability.Note that a collection of uniform distributions on a sequence of sets S1,S2,...,where each S k is a(τ(k),ε(k))-pseudorandom set,constitutes a pseudorandom ensemble,pro-vided that both functionsτ(n)andε−1(n)grow faster than any polynomial.Therefore,we shall refer to such a sequence of sets as a pseudorandom ensemble.We now present the concept of"evasive sets".Informally,evasiveness means that it is hard,for efficient algorithms,tofind strings which belong to these sets.Definition3.2:Let S1,S2,...be a sequence of(non-empty)sets such that for every n, S n⊆{0,1}Q(n),for afixed polynomial Q.Such a sequence is called a polynomially-evasive(denoted P-evasive)ensemble if for any probabilistic polynomial-time algorithm A,any constant c,sufficiently large n,and any x∈{0,1}nProb(A(x)∈S n)<n−cwhere the probability is taken over the random coins of algorithm A.The following theorem states the existence of a P-evasive ensemble which is also pseu-dorandom.Theorem3.1[GK]:There exists a P-evasive pseudorandom ensemble S1,S2,...with Q(n)=4n.Furthermore,there exists a Turing machine which on input1n outputs the set S n.For disproving the parallel composition theorem we shall need a stronger notion of ly,one which resists also non-uniform algorithms.This definition of evasiveness involves a collection of sets for each length,rather than a single set per length as in the uniform case.Definition3.3:Let Q(.)be a polynomial,and for n=1,2,...let S(n)be a collection of2n sets{S1(n),...,S2n(n)},where each S i(n)⊆{0,1}Q(n).The sequence S(1),S(2),...is called a non-uniform polynomially evasive(denoted P/poly-evasive)ensemble if for any c>0,for sufficiently large n and any(probabilistic)circuit C of size n c(with n inputs and Q(n)out-puts)1Prob(C(i)∈S i)<n cwhere the probability is taken over the random coins of C and i∈{1,...,2n},both with uniform probability.That is,a sequence S(1),S(2),...is a P/poly-evasive ensemble if any circuit of size polynomial in n,which gets a randomly selected index of one of the sets in S(n),cannot succeed to output an element in that set,except for a negligible probability.Remark:Notice that in the definition of P-evasive ensembles the(uniform)algorithm trying to hit an element in the evasive set S n,gets as input a string x of length n,which can be seen as an auxiliary input.The crucial difference between this"uniform" definition and the definition of P/poly-evasiveness is that in the latter the auxiliary input is allowed to be of any length polynomial in the length of the target strings,while in the former the auxiliary input is properly shorter than the target strings in the set S n.The following theorem states the existence of a P/poly-evasive ensemble which is composed of pseudorandom sets.Theorem3.2:There exists a P/poly-evasive ensemble S(1),S(2),...with Q(n)=4n,such that for every n,each S i(n)is a(2n/4,2−n/4)-pseudorandom set of cardinality2n.Further-more,there exists a Turing machine which on input1n outputs the collection S(n).The proof of this theorem is given in the Appendix.4.SEQUENTIAL COMPOSITION OF ZERO-KNOWLEDGE PROTOCOLSA natural requirement from the notion of zero-knowledge proofs is that the informa-tion obtained by the verifier during the execution of a zero-knowledge protocol will not enable him to extract any additional knowledge from subsequent executions of the same protocol.That is,it would be desirable that the sequential composition of zero-knowledge protocols would yield a protocol which is itself zero-knowledge.Such a pro-perty is crucial for applications of zero-knowledge protocols in cryptography(for details and further motivation see[GO,Ore]).We prove that the original definition of(computational)zero-knowledge introduced by Goldwasser,Micali and Rackoff in[GMR1]is not closed under sequential composi-tion.Several authors have previously observed that this definition probably does not guarantee the robustness of zero-knowledge under sequential composition,and hence have introduced more robust formulations of zero-knowledge[Fei,GMR2,GO,Ore, TW].But so far,no proof has been given for the claim that computational zero-knowledge(with uniform verifiers)fails sequential composition.Intuitively,the reason that a zero-knowledge protocol could not be closed under sequential composition is that the definition of zero-knowledge requires that the informa-tion transmitted in the execution of the protocol is"useless"for any polynomial-time computation;it does not rule out the possibility that a cheating verifier could take advan-tage of this"knowledge"in a subsequent interaction with the(non-polynomial time) prover for obtaining valuable information.This intuition(presented in[Fei])is the basis of our example of a protocol which is zero-knowledge in a single execution but reveals significant information when composed twice in a sequence.This protocol,presented in the proof of the following theorem,uses a P-evasive ensemble as defined in Definition 3.2and whose existence is stated in Theorem3.1.Theorem4.1:Computational Zero-Knowledge([GMR1]formulation)is not closed under sequential composition.Proof:Let S1,S2,...be a P-evasive pseudorandom ensemble as described in Theorem 3.1.Also,let K be an(arbitrary)hard Boolean function,in the sense that the language L K={x:K(x)=1}is not in BPP(we use this function as a"knowledge"function).。
高中语法学案--特殊名词形式变化背记清单(附练习答案)
特殊名词变化背记清单一、名词的特殊复数形式：(一) 名词的复数形式不规则变化：1. child ---children小孩2. foot --- feet脚3.toot--- teeth牙齿4. goose --- geese鹅5. mouse ---mice老鼠6. ox ---oxen 公牛7. analysis --- analyses分析8. crisis --- crises危机9. thesis --- theses论文10. man --- men男人11. woman --- women女人12. medium --- media 媒体13.basis --- bases基础14. datum --- data数据15. bacterium --- bacteria细菌16. criterion --- criteria标准17. phenomenon --- phenomena现象18. German --- Germans 德国人19. Frenchman--- Frenchmen 法国人20. Englishman ---Englishmen 英国人(二) 有些名词单复数同型:1. Chinese --- Chinese中国人2. Japanese --- Japanese日本人3. Swiss--- Swiss瑞士人4. spacecraft --- spacecraft 航天器5. aircraft --- aircraft 飞机6.headquarters---headquarters 总部7. means ---means 方式8. species --- species 物种9. news --- news 消息10. deer --- deer 鹿11. fish --- fish鱼12. sheep --- sheep 绵羊(三) 有些以f结尾的名词变复数直接加s:1.belief --- beliefs 信念2.roof --- roofs 屋顶3.cliff --- cliffs 悬崖4.staff --- staffs 员工5.proof --- proofs 证据6.chief --- chiefs 首领7.gulf --- gulfs 海湾8.grief --- griefs 伤心事9.serf --- serfs 农奴10.safe --- safes 保险箱(四) “单位”名词的单复数：1. dollar--- dollars美元2. franc--- francs法郎3. yuan --- yuan 元4. meter/metre --- meters 米5. jin --- jin 斤6. mu --- mu 母(五) 复合名词的复数形式：1. looker-on --- lookers-on旁观者2. passer-by --- passers-by过路人3. man doctor --- men doctors男医生4. woman teacher -- women teachers 女老师5 brother-in-law -- brothers-in -law 姐夫（妹夫）6. sister-in-law --- sisters-in-law 嫂子（弟媳）7. mother-in-law --- mothers-in-law 岳母（婆婆）8. father-in-law --- fathers-in-law 岳父（公公二、下列名词没有单数形式, 是复数概念，在句中作主语时谓语动词用复数形式：1. clothes2. police3. cattle4. people三、下列名词的单复数含义不同：1. paper纸--- papers试卷，文件，论文2. good 好的--- goods 货物3. ash 灰--- ashes 骨灰，遗骸4. glass 玻璃--- glasses 眼镜5. sand 沙子--- sands 沙滩6. wood 木材--- woods 树林7. green绿色的---greens 青菜8. time 时间--- times 时代9. drink酒--- drinks 饮料10. arm 手臂--- arms 武器11. look看---looks 外表，面容12. line 线--- lines 台词13. manner方式,样式---manners 礼貌,规矩14. work 工作--- works 工厂,作品,工事15. damage 损害---damages 损失赔偿费,费用16. cloth 布--- clothes 衣服17. pain 疼痛--- pains 辛劳，痛苦18. ruin 毁坏--- ruins 废墟19. fruit 水果--- fruits 各种水果20. noise 噪音--- noises 各种噪音21. color 颜色--- colors 各种颜色22. custom 风俗;习惯--- customs 海关23. fish 鱼肉--- a fish鱼24. chicken 鸡肉--- a chicken 小鸡25. tin 锡--- a tin 罐头26. relation 关系--- a relation 亲戚,亲属27. iron 铁--- an iron 熨斗28. beauty 美丽--- a beauty 美人,美的东西29. power 威力;力量;电力--- a power 大国30. room空间--- a room 房间31. glass 玻璃--- a glass 玻璃杯32. success成功--- a success 一个成功的人(事)33. communication 通讯;交流--- communications 通讯系统,通讯工具34. convenience 方便，便利--- conveniences便利设备35. necessity 需要,必要性--- necessities 必需品四、构词法----名词和动词的相互转化：1. 很多动词可以转化为名词：1) have a look ( chat, talk, wash, swim, rest, try, quarrel, interview, taste, slip, ect.)2) make a study ( guess, call, survey, change, answer, slip, visit, appointment ect.)3) come to a stop (a pause, an end, ect.)2. 有些名词也可当动词用：1) Have you booked your ticket? (预订)2) The hall can seat two thousand people. (容纳)3) We’ll back you up. (支持)4) We’ll head for Shanghai tomorrow. (朝….前进)5) If so, you will be badly fooled. (上当,愚弄)6) They were hosted by the members of the embassy. (招待)7) This helped to bridge over our difficulties. (度过)8) Over three hundred students stormed into the building. (涌入；冲进)五、注意下列动词形容词变化的名词：1. grow growth2. warm warmth3. true truth4. repeat repetition5. describe description6. argue argumentpete competition8.necessary necessity9.recognize recognition10.decide decision11.confuse confusion12.divide division13.relax relaxation14.pronounce pronunciation15.propose proposal16.arrive arrival 17.refuse refusal18.survive survival19.remove removal20.approve approval21.proud pride22.dry drought23.permit permission24.discover discovery25.speak speech26.fortunate fortune27.differ difference28.strong strength29.marry marriagerm information31.popular popularity32.succeed success六、用名词的正确形式填空:1.There are two ____________ (box) of ____________ (match), three ____________ (knife), four ____________(brush), five ____________ (key), six ____________ (photo), seven ____________ (dictionary), eight____________ (tomato), nine ____________ (glass), ten ____________ (radio) and eleven ____________ (toy bus) on that table.2.Look at those ____________ (wolf), ____________ (sheep), ____________ (deer), ____________ (fox),____________(ox) ____________ (monkey), and ____________ (goose).3.Who are they? They are all ____________ (hero). And they are all ________________ (woman doctor)4.There are so many public____________ (convenience) in cities .5.Three ____________ (Swiss), two ____________(German) and five _____________(Frenchman) are present at themeeting.6.Those little _____________(child) are playing with those little____________ (mouse).7.The rainbow is one of the most beautiful ____________ (phenomenon) in nature.8.Two____________(aircraft) and two__________(spacecraft) are being designed in the ___________(headquarters).9.We have tried all the ________(means), but not every_________(means) has worked.10.It tastes like________(chicken). Do you raise ________ (chicken)?11.He regarded music as one of life's ____________ (necessary).12.They took great __________ (proud) in their excellent children.13.English __________________ (pronounce) is terribly difficult.14.The job involved the constant _____________ (repeat) of the same movements15.The police have issued a detailed ______________ (describe) of the missing woman.16.The ____________ (consume) of alcohol on the premises（经营场所）is forbidden.17.Sometimes there's a lot of ____________ (compete) between children for their mother's attention.18.There is general_________ (recognize) that the study techniques of many students are weak.19.The news aroused a lot of _____________ (curious) among local people.20.The fear of unemployment can be a source of deep ___________ (anxious) to people.21.I think the government has lost touch with___________ (real).22.The lake has more than 20 _______________ (vary) of fish.23.We need an effective strategy to fight ____________ (poor).24.This latest interview was further __________ (prove) of how good at her job Cara was.25.Several members hold very right-wing _________(believe) .26.This year (a) severe ___________ (dry) has ruined the crops.27.With seven people squashed（塞入）in one house, you don't get much ___________ (private).28.In most people’s eyes, she was nothing more tha n a common ___________ (crime).29.It was his ___________ (brave) that saved the child.30.Meditation allows you to enter a state of deep_____________(relax) .31.The thought occurred to me that he might not be telling the _________(true).32.All at once they were aware of the evening of light and___________ (warm).33.We finally came to a firm _________ (decide) on the matter.34.The French government has approved ___________ (propose) for a new waste law.35.The party hopes to win the _____________ (argue) about how to reform the health systemkey名词key: 1. boxes, matches, knives, brushes, keys, photos, dictionaries, tomatoes, glasses, radios, buses2. wolves, sheep, deer, foxes, oxen, monkeys, geese3. heroes, women doctors4. conveniences5. Swiss, Germans, Frenchmen6. children, mice7. phenomena 8. aircraft, spacecraft, headquarters9. means, means 10. chicken, chickens11. necessities 12. pride 13. pronunciation 14. repetition 15. description 16. consumption 17. competition 18. recognition 19. curiosity 20. anxiety 21. reality 22. varieties23. poverty 24. proof 25. beliefs 26. drought 27. privacy 28. criminal 29. bravery 30. relaxation 31. truth 32. warmth 33. decision 34. proposals 35. argument。
机器证明-埃尔多斯-谢卡雷斯问题说明书
Mechanical Proving for ERDÖS-SZEKERES Problem1Meijing ShanInstitute of Information science and Technology,East China University of Political Science and Law, Shanghai, China. 201620Keywords: Erdös-Szekeres problem, Automated deduction, Mechanical provingAbstract：The Erdös-Szekeres problem was an open unsolved problem in computational geometry and related fields from 1935. Many results about it have been shown. The main concern of this paper is not only show how to prove this problem with automated deduction methods and tools but also contribute to the significance of automated theorem proving in mathematics using advanced computing technology. The present case is engaged in contributing to prove or disprove this conjecture and then solve this problem. The key advantage of our method is to utilize the mechanical proving instead of the traditional proof and this method could improve the arithmetic efficiency.IntroductionThe following famous problem has attracted more and more attention of many mathematicians [3, 6, 12, 16] due to its beauty and elementary character. Finding the exact value of N(n) turns out to be a very challenging problem. The problem is very easy to explain and understand.The Erdös-Szekeres Problem 1.1 [4, 15]. For any integer n ≥ 3, determine the smallest positive integer N(n) such that any set of at least N(n) points in generalposition in the plane contains n points that are the vertices of a convex n-gon.A set of points in the plane is said to be in the general position if it contains no three points on a line. This problem was also called Happy Ending Problem by Erdös, because two investigators Esther Klein and George Szekeres who first worked on the problem became engaged and subsequently married[8, 17].The interest of Erdös and Szekeres in this problem was initiated by Esther Klein(later Mrs. Szekeres), who observed that from 5 points of the plane of which no three lie on the same straight line it is always possible to select 4 points determining convex quadrilateral. There are three distinct types of five points in the plane, as shown in Figure 1.Figure 1. Three cases for 5 points.In any case of the Figure 1, one can find at least one convex quadrilateral determined by the points. Klein [4] suggested the following more general problem: Can we find for a given n a number N(n) such that from any set containing at least N points it is possible to select n points forming a convex polygon?As observed by Erdös and Szederes [4], there are two particular questions:(1) Does the number N corresponding to n exist?(2) If so, how is the least N(n) determined as a function of n?1T his work was financially supported by Humanity and Social Science Youth foundation of Ministry of Education of China (No. 14YJCZH020).They proved the existence of the number N(n) by two different methods. The first one is a combinatorial theorem of Ramsey. The foundation of the second one isbased on some geometrical and combinatorial considerations. And then they formulated the following conjecture.Conjecture 1.1 N (n ) = 2n −2 + 1 for all n ≥ 3.Despite its elementary characters and the efforts of many researchers, the ErdösSzekeres problem is solved for the value n = 3, 4 and 5 only. The case n = 3 istrivial, and n = 4 is due to Klein. The equality N (5) = 9 was proved by E. Makaiwhile the published proof by Kalbfleisch [11] and then Bonnice [2] and Lovasz [13]independently published the much simpler proofs. The bottle neck of this problem now is when n >5, how to prove or disprove the conjecture.About this problem, the best currently known bounds are(1)Where is a binomial coefficient. The lower bound was obtained by Erdös andSzekeres [4] and the upper bound is due to Töth and Valtr [18]. The lower bound issupposed to be sharp, according to conjecture 1.1.In this paper we use the automated deduction method and tools [1, 19-21] to establish a mechanical method for proving problem instead of the manual proof. We hope the method might give a rise to substantially promote this unsolved problem.The rest content of the paper is indicated by section headings as follows: Section 2 preliminaries, Section 3 main results, Section 4 conclusion and remarks.PreliminariesIn this Section, we present some algorithms that would help us develop our method in next section.Algorithm 2.1. Modified Cylindrical Algebraic Decomposition (CAD). Due to the problem statement, the proof should consider all kinds of points’ positions on the plane.The Cylindrical Algebraic Decomposition [5, 14] of R n adapted to a set ofpolynomials which is a partition of R d is cells (simple connected subsets of R d )such that each input polynomial has a constant sign on each cell. Basically, thealgorithm computes recursively at least one point in each cell (so that one can testthe cells that verify a fixed sign condition). The sample points Pi got by originalCAD are always more than one on each cell. We modify the procedure to have asample point on each cell of the final cell decomposition, by the rule that Pi has a constant sign on each cell. We elaborate the main idea under lying our method by showing how our main algorithm evolved from the original one.Here, we describe it as follows.Algorithm. MCADInput: A set F of polynomialsOutput: A F-sign-invariant CAD of R nStep 1. Projection. Compute the projection polynomials which using exclusivelyoperations, and receive some (n − 1) -variate polynomials.Step 2. Recur. Apply the algorithm recursively to compute a CAD of R n −1which Q(F )is sign-invariant.Step 3. Lifting. Lift the Q(F )-sign-invariant CAD of R n −1 up to a F-sign-invariant CAD of using the auxiliary polynomial Π(F ) of degree no largerthan d (F ) (d is the maximum degree of any polynomial in F).Step 4. Choice. Utilize the strategy that {Pi} has constant sign on each cell tochoose one sample point on each cell.Algorithm 2.2 (Graham Scan Algorithm) [9, 10, 22]We present one of the simplest algorithms used to find the convex hull fromsome points. Some basic definitions are provided in the field of Computational Geometry. This algorithm works in three phases:Input: A set S of pointsOutput: The convex hull of S.Step 1. Find an extreme point. The algorithm starts by picking a point in S known to be a vertex of the convex hull. This point is chosen to be with smallest y coordinate and guaranteed to be on the hull. If there are some points with the same smallest y coordinate, we will choose the point with largest x coordinate in them. In other words, we select the right most lowest point as the extreme point.Step 2. Sort the points. Having selected the base point which is called P0 , then the algorithm sorts the other points P in S by the increasing counter-clockwise (ccw) angle the line segment P0P makes with the x-axis. If there is a tie and two points have the same angle, discard the one that is closest to P0 .Step 3. Construct the convex hull. Build the hull by marching around the star shaped polygon, adding edges when we make a left turn, and back-tracking when we make a right turn. We end up with a star-shaped polygon, see Figure 3 (one in whichone special point, in this case the pivot, can “see” the whole polygon). Considering efficiency in Step 2, it is important to note that the comparison of sorting between two points P2 and P3 can be made without actually computing their angles. In fact, computing angles would use slow in accurate trigonometry functions, and doing these computations would be a bad mistake. Instead, one just observes that P2 would make a greater angle than P1 if (and only if) P2 lies on theleft side of the directed line segment P0P1 , see Figure 2.We make full use of this algorithm to judge whether the polygon received in every recursive step is a convex polygon or not. It is a decision method in our algorithm.To state the algorithm clearly, we will describe it in a style of pseudo-code.Algorithm. Graham Scan AlgorithmInput: A set S of points in the planeOutput: A list containing the vertex of the convex hullSelect the right most lowest point P0 in SFigure 3. Graham ScanFigure 2. Sort the points.Sort S angularly about P0 as a center.For ties, discard the closer points.Let S be the sorted array of points.Push S [1] = P0 and P1 onto a stack Ω.Let P1 = the top point on ΩLet P2 = the second top point on Ωwhile S [k ] P doif (S [k −1] is strictly left of the line S [k ] to S [k +1]), thenPush S [k −1] onto ΩelsePop the top point S [k ] off the stack Ω.fi;od;Main ResultsIn contrast with the traditional proof, this method presented following can show the convex polygons received in every step.The main idea of our algorithm: First, give randomly four points in general position in the plane, and then we use polynomials of points’ coordinates to represent the lines. We extend the 4-element set to some5-element sets. We do this by establishing the corresponding Modified Cylindrical Algebraic Decomposition (MCAD) and design an interactive program which allow the user to choose among the candidate one sample point in each cell. We use Graham Scan algorithm to determine whether or not there is a convex 5-gon (convexpentagon) in every set received. If some of the received 5-element sets have no convex 5-gon, then extend them to the 6-element sets by the strategy mentionedabove. Simultaneously, check whether or not each of the received 6-element sets has any convex hull at least with any convex 5-gon in the polygon. We implement theprogram repeatedly until can find a convex n-gon (n ≥ 5) in any set. We trace the processing of the extension and decision, and then draw a conclusion that N (5) = 9.To prove this approach, we write the following algorithm named “conv5”. Based on this algorithm can generate short and readable mechanical proving for the Erdös-Szekeres conjecture, including the case of n = 3, 4, 5. It consists of two main algorithms—Modified Cylindrical Algebraic Decomposition algorithm and Graham Scan algorithm and some sub-algorithms such as collinear, pol, sam, ponlist, min0,isleft, ord, conhull, point5, convex, G5,G6,Pmn.Algorithm Conv5Input: four points in the general position in the plane.Output: Any polygon with at least 9 points in general position in the planecontains a convex 5-gon. Step 1 [collinear]. Write the line polynomials with the given four points (basepoints).Step 2 [pol, sam, ponlist]. Illustrate how we utilize the CAD to find somesample points in the cell which built by the lines, and then with the base n points get n +1 -element sets.Step 3 [min0, isleft, conhull, point5, convex, G5]. Decide whether or not thereis a convex hull or a convex n-gon (n ≥ 5) in every set; if it is true, then stop; elsegoto Step 4.Step 4 [G6, Pmn]. Deal with the sets which have no convex n-gon (n ≥ 5).Recursively implement Step 2 and Step 3 process, until at least there is a convex 5-gon in any set. End Conv5.The key techniques of the algorithm are listed as follows:1. To reduce the complexity of the computation and increase the efficiency,when we check whether or not there is a convex 5-gon in the given points set,we utilize the strategy as follows:if there is a convex hull at least with 5 points in the points set thenpop this points setelif there is a convex 5-gon thenpop this points setelse “there is no convex 5-gon in this points set” go to next step2. Each convex n-gon (n ≥ 5) contains a convex 5-gon3. If there is no convex 4-gon, then there should be no convex 5-gon.Conclusion and RemarksBy the Maple procedure we have implemented the mechanical method for the conjecture in certain cases. Through observing the whole computational process, we obtain a certain answer that any set with at least 9 points in general position in the plane contains a convex 5-gon. This method can be generalized in an obvious way to arbitrary base points in the plane.For the mechanical method proposed here, on one hand it provided one of the promising direction for proving or disproving the Conjecture 1.1 (when n ≥ 6), even for handling with some unsolved problems in computational geometry. On the other hand, it gave one especially useful application of computer algebraic andautomated deduction. For further investigations, now we consider about the following problems:1. Does any set of at least 17 points in general position in the plane contains 6 points which are the vertices of a convex hexagon? Can we give the proof about N (6) existence and prove or disprovethe corresponding conclusion by mechanical proving? Now the best known conclusion about this is N (6) ≥ 27, if it exists.2. Erdös posed a similar problem on empty convex polygons. Whether or not we can give the automated proof to this problem?References[1] M. de Berg, M. van Kreveld, M. Overmars and O. Schwarzkopf, omputational GeometryAlgorithm and Applications, (2nd ed.), Spring-Verlag, Berlin, Heideberg, New York, 1997.[2] W. E. Bonnice, On convex polygons determined by a finie planar set, Amer. Math. Monthly.[3] F. R. K. Chung and R. L. Graham, Forced Convex n-gons in the Plane, Discr. Comput. Geom. 19(1998), 367-371.[4] P. Erdös and G. Szekeres, A combinatiorial problem in geometry, Comositio Mathematica 2(1935), 463-470.[5] E. Collins George, Quantifier elimination for the elementary theory of real closed fields by cylindrical algebraic decomposition Lecture Notes In Computer Sciencevol. 33, Springer-Verlag, Berlin, pp. 134-183.[6] Kráolyi Gyula, An Erdös-Szekeres type problem in the plane.[7] X. R. Hou and Z. B. Zeng, An efficient Algorithm for Finding Sample Points of Algebraic Decomposition of Plane, Computer Application, 1997, (in Chinese).[8] P. Hofiman, The Man Who loved Only Numbers Hyperion, New York, 1998.[9] /ah/alganim/version0/Graham.html.[10] http://cgm.cs.mcgill.ca/ beezer/cs507/main.html.[11] J. D. Kalbfleisch, J. G. Kalbfleisch and R. G. Stanton, A combinatorial problem on convexregions, Proc. Louisiana Conf. Combinatorics, Graph Theory and Computing, Louisianna StateUniv., Baton Touge, La, Congr. Numer. 1 (1970), 180-188.[12] D. Kleitman and L. Pachter, Finding convex sets among points in the plane, Discr. Comput.Geom. 19, (1998), 405-410.[13] L. Lovasz, Combinatorial Problem and Exercises North-Holland, msterdam, 1979.[14] Bhubaneswar Mishra, Algorithmic Algebra, Springer-Verlag, 2001.[15] W. Morris and V. Soltan, The Erdös-Szekeres Problem on Points in ConvexPosition- A, Survey Bulletin of the American Mathematical Society, vol. 37.[16] L. Graham Ronald and Yao Frances, Finding the Convex Hull of a SimplePolygon Report No. STAN-CS-81-887, 1998.[17] B. Schechter, My Brain is Open Simon Schuster, New York, 1998.[18] G. Tóth and P. Valtr, Note on the Erdös-Szekeres theorem, Discr. Comput. Geom. 19 (1998), 457-459.[19] W. T. Wu, Basic Principles of Mechanical Theorem Proving in Geometries Science Press,Beijing, 1984, (Part on elementary geometries, in Chinese).[20] L. Yang and B. C. Xia, Automated Deduction in Real Geometry, Geometric Computation, WorldScientific, 2004.[21] L. Yang and Z. Z. Jing and X. R. Hou, Nonlinear Algebraic Equation System and AutomatedTheorem Proving, Shanghai Scientific and Technological Education Publishing House, Shanghai,1996, (in Chinese).[22] P. D. Zhou, Computational Geometry Design and Analysis in Chinese, TsingHua UniversityPress, 2005.。
离散数学及其应用重要名词中英对应以及重要概念解释与举例
离散数学及其应用重要名词中英对应以及重要概念解释与举例1 The Foundations: Logic and Proofs（逻辑与证明）1.1 Propositional Logic（命题逻辑）Propositions（命题）——declarative sentence that is either true or false, but not both.判断性语句，正确性唯一。
Truth Table（真值表）Conjunction（合取，“与”，and），Disjunction（析取，or，“相容或”），Exclusive（异或），Negation（非，not），Biconditional（双条件，双向，if and only if）Translating English Sentences1.2 Propositional Equivalences（命题等价）Tautology（永真式、重言式），Contradiction（永假式、矛盾式），Contingency（偶然式）Logical Equivalences（逻辑等价）——Compound propositions that have the same truth values in all possible cases are called logical equivalent.（真值表相同的式子，p<->q是重言式）Logical Equivalences——Page24Disjunctive normal form(DNF，析取范式)Conjunctive normal form(CNF，合取范式) 见Page27～291.3 Predicates and Quantifiers（谓词和量词）Predicates——谓词，说明关系、特征的修饰词Quantifiers——量词? Universal Quantifier(全称量词) "全部满足? Existential Quantifier(存在量词) $至少有一个Binding Variables(变量绑定，量词作用域与重名的问题)Logical Equivalence Involving QuantifiersNegating Quantified Expressions(量词否定表达：否定全称=存在否定，否定存在=全程否定) Translating from English into Logical Expressions(自然语句转化为逻辑表达)Using Quantifiers in System SpecificationsExamples from Lewis Carrol——全称量词与条件式(p->q)搭配，存在量词与合取式搭配。
2025版《师说》高中全程复习构想英语(北师大版)第三讲 名词和数词
4．[2022·新高考Ⅰ卷]The GPNP's main goal is to improve connectivity between separate ________ (population)and homes of giant pandas, and eventually achieve a desired level of population in the wild.
答案与解析：photographer 考查名词。此处表示John Olson之前是一名摄影师。 分析句子结构可知，空格前有冠词“a”和形容词“former”修饰，此处应用可数名词 photographer的单数形式。故填photographer。
7．[2022·全国乙卷]As a main promoter of the International Tea Day, the birthplace of tea and the largest teaproducing country, China has a ________ (responsible) to work with other countries to promote the healthy development of the tea industry.
答案与解析：warning 考查名词。形容词weighty后接名词形式。根据句中不 定冠词a可知，用单数形式。
3．[2023·全国乙卷]But for all its ancient buildings, Beijing is also a place that welcomes the fastpaced development of modern life, with 21stcentury architectural ________ (wonder) standing side by side with historical buildings of the past.
信息安全ppt
Hybrid Crypto
• Diffe-Hellman key exchange
• Alice and Bob agree on a finite cyclic group G (Multiplicative group of integers mod p)
▫ Period p, prime number ▫ Base g, primitive root mod p
• Alice picks a random natural number a and sends ga mod p to Bob.
• Bob picks a random natural number b and sends gb mod p to Alice.
• Alice computes (gb mod p)a mod p • Bob computes (ga mod p)b mod p • Both know gab mod p = gba mod p
• Preparation
▫ Alice prepares a state, sends to Bob, measures
• Entanglement
▫ Alice and Bob each receive half the pair, measure
Non-Orthogonal Bases
• Complementary bases
exchange ▫ Careful adherence to process
Symmetric Key Crypto
• The same (or similar) key
▫ For both encryption and decryption
微观经济学理论部分英文词汇
46, sequences
n. [数][计] 序列，顺序；继起的事（sequence的复数形式）
Sequences: 序列 | 图形逻辑 | 数列
47, semester [sɪ'mestə]
n. 学期；半年
48, relatively ['relətɪvlɪ]
Reunited: 重遇 | 重聚 | 团聚
9, proofs [pru:fs]
n. 证明；证据（proof的复数）；校稿
10, proceeds ['prəʊsiːdz]
n. 收入，收益；实收款项
11, extension [ɪk'stenʃ(ə)n; ek-]
n. 延长；延期；扩大；伸展；电话分机
n. 筛选；放映；[物] 屏蔽；审查；防波
v. 筛选；拍摄（screen的ing形式）；遮蔽；隔挡
Screening: 场次 | 筛选 | 筛查
56, mechanism design
机制设计（微观经济学和博弈论的分支领域）；机构设计（学科名称）；机械设计
57, altered ['ɔːltəd]
27, Fixed Point Theorems
Fixed Point Theorems: 不动点定理
28, coverage ['kʌv(ə)rɪdʒ]
n. 覆盖，覆盖范围
coverage: 新闻报道 | 覆盖面 | 保险范围
29, instructors
n. 教练；讲师；指导书（instructor的复数）
adj. 人际的；人与人之间的
离散数学及其应用重要名词中英对应以及重要概念解释与举例
离散数学及其应用重要名词中英对应以及重要概念解释与举例1 The Foundations: Logic and Proofs（逻辑与证明）1.1 Propositional Logic（命题逻辑）Propositions（命题）——declarative sentence that is either true or false, but not both.判断性语句，正确性唯一。
Truth Table（真值表）Conjunction（合取，“与”，and），Disjunction（析取，or，“相容或”），Exclusive（异或），Negation（非，not），Biconditional（双条件，双向，if and only if）Translating English Sentences1.2 Propositional Equivalences（命题等价）Tautology（永真式、重言式），Contradiction（永假式、矛盾式），Contingency（偶然式）Logical Equivalences（逻辑等价）——Compound propositions that have the same truth values in all possible cases are called logical equivalent.（真值表相同的式子，p<->q是重言式）Logical Equivalences——Page24Disjunctive normal form(DNF，析取范式)Conjunctive normal form(CNF，合取范式) 见Page27～291.3 Predicates and Quantifiers（谓词和量词）Predicates——谓词，说明关系、特征的修饰词Quantifiers——量词Ø Universal Quantifier(全称量词) "全部满足Ø Existential Quantifier(存在量词) $至少有一个Binding Variables(变量绑定，量词作用域与重名的问题)Logical Equivalence Involving QuantifiersNegating Quantified Expressions(量词否定表达：否定全称=存在否定，否定存在=全程否定) Translating from English into Logical Expressions(自然语句转化为逻辑表达)Using Quantifiers in System SpecificationsExamples from Lewis Carrol——全称量词与条件式(p->q)搭配，存在量词与合取式搭配。
北京邮电大学 计算机学院 离散数学 1.7&1.8- proof strategy
Each of the implications Pi Q is a case. You must
Convince the reader that the cases are inclusive, i.e., they exhaust all possibilities establish all implications
1Q
The Contrapositive of this assertion is
from which it follows that Q must be true.
2015-2-6
College of Computer Science & Technology, BUPT
13
Example
2015-2-6
College of Computer Science & Technology, BUPT
4
Direct proof
assumes the hypotheses are true uses the rules of inference, axioms and any logical equivalences to establish the truth of the conclusion.
Discrete Mathematical Structures
Methods of Proof （证明的方法）
Yang Juan
yangjuan@
College of Computer Science & Technology Beijing University of Posts & Telecommunications
proof的修改问题（ZHUAN）
proof的修改问题（ZHUAN）当通讯作者收到⽂章的proof（校样）后，可以修改作者信息、联系⽅式、增添作者、调换作者顺序、添加基⾦等，如果删除作者、调换作者，尤其是第⼀作者和通讯作者时，⼤多期刊编辑会要求提供变更说明并签字的证明。
这⾥不详述了。
随proof附件的，⼤多期刊还有query 和annotate⽂件（有的期刊没有annotate⽂件）Proof还有内容的修改，包括期刊编辑校对时给你的pdf标注，有的期刊没有pdf标注，直接将排版好的proof发给你校对。
不管怎样，作者可以在编辑发给你的proof中直接进⾏修改错误的内容，⼀定要注意按照proof修改的要求进⾏删除、增添、批注等操作。
没有annotate⽂件的，就按照pdf⼯具栏⾥的各个图标所表⽰的意思或功能进⾏操作，在proof中，为了区别排版编辑对你稿件的修改，作者修改proof时除了删除、增添、批注等操作外，还可以对这些删除、增添、批注等进⾏黄⾊加亮显⽰。
修改好后，在⽂档⽂件名后加上“_corrected”字样，然后将附件回复到编辑邮箱，记得要提醒编辑收到后给你回信。
Query⽂档⾥，response的内容可以在下⽅填上（针对query，⼀条条列出修改页数和⾏数、修改内容和备注说明）；如果已经在proof中修改了（修改信息对Query进⾏了回答），可以在response⾥指出修改页和⾏数、备注说明。
Query⽂档写好后打印传真给指定的传真号。
同样，记得提醒编辑给你回信。
本帖附件1：INSTRUCTIONS ON THE ANNOTATION OF PDF FILES （如何对pdf⽂档进⾏批注、修改等说明）本帖附件2：AUTHOR QUERY FORM本帖附件3：MAMT_1528_corrected （黄⾊加亮部分是作者的修改，其它地⽅是排版编辑的修改。
）本⽂已经正式出版。
⽅东明北京⼤学微电⼦学研究院附上proof时的往返email信件Dear Sir/Madam,The attached document is the corrected proof. All corrections are highlighted in yellow. Please give me a reply when you receive the attachment. Thank you very much.Best Regards,Dongming Fang2009-01-12________________________________________________________________________________发件⼈： Elsevier Ltd, Editorial-Production Department发送时间： 2009-01-09 22:45:09收件⼈：抄送：主题： [!! SPAM] Proofs of [MAMT_1528]P-authorproof-v11(a)Dear AuthorThe proof of your article, to be published in Mechanism and Machine Theory, is attached to this e-mail as a PDF file. Also attached are instructions on the annotation of PDF files. A `Query Form' may also be attached if any questions regarding your article have arisen during the preparation of the proof.We will do everything possible to get your article published quickly and accurately; to do this we need your cooperation. Please respond promptly, even if you have no corrections: the sooner we hear from you, the sooner your corrected article will appear online. Please note that any delay in returning your corrections could result in a delay in publication and that any significant changes to the article as accepted for publication will only be considered at this stage with the permission of the Editor.Please note that proof corrections can now be annotated on-screen, which allows you to mark directly in the PDF file, and return the marked file as an e-mail attachment. See the attached instructions for further information.Alternative methods of returning proof corrections:If you do not wish to use the PDF annotations function, you may list the corrections (including replies to the Query Form) in an e-mail and return them to us using the `reply' button to this e-mail. Please list your corrections quoting line number.If, for any reason, this is not possible, mark the corrections and any other comments (including replies to the Query Form) on a printout of your proof and fax this to the number given below, or scan the pages and return them by e-mail.We prefer to receive your corrections by e-mail or fax so that we can process your article quickly and efficiently. However, if you wish to return your corrections by post then please contact us and we will provide the full postal address.Please use this proof for checking the typesetting, editing, completeness and correctness of the text, tables and figures. To assist you withthis, copy-editing changes have been highlighted in red with the original text in a pop-up box.If you submitted usable colour figures with your article they will appear, at no extra charge, in colour on the web if reproduced in colour in the attached PDF proof of your article. In the printed issue, colour reproduction depends on journal policy and whether or not you agree to bear any costs (not applicable to journals which appear only online). Any `supplementary' material to your article (i.e., not appearing in print) will be accessible after your corrected article is placed online; such material is not part of the proofing procedure and is therefore not attached here.Before returning your proof corrections: please ensure that you have answered any questions raised on the Query Form and that you have indicated all corrections: the processing of any subsequent corrections cannot be guaranteed.Kind regards,ElsevierE-mail corrections to:Fax: +31 2048 52799。
顺口溜巧记以f或fe结尾的名词变复数
顺口溜巧记以f或fe结尾的名词变复数一个小偷(thief)的妻子(wife),用树叶(leaf)做的小刀(knife),杀死了一只狼(wolf),把它劈成两半(half),挂在衣架上.以f,fe结尾的名词,变为复数时,要变f,fe为ves!妻子持刀去宰狼，小偷吓得发了慌;躲在架后保己命，半片树叶遮目光。
按：顺口溜中的黑体字是中学阶段学过的九个以—f(e)结尾的名词：wife(妻子)，knife(小刀)，wolf(狼)，thief(小偷)，shelf(架子)，self(自己)，life(生命)，half(一半)，leaf(树叶)。
这九个词变复数时，都是改-f(e)为ve再加-s。
由self构成的复合词，其变化与self相同(如：myself→ourselves;yourself→yourselves;himself，herselfitself→themselves)。
顺口溜巧记以f或fe结尾的名词变复数一、变f或fe为v再加-es的树叶半数自己黄，妻子拿刀去割粮；架后窜出一只狼，就像小偷逃命忙。
【解释】在这些词中，由单数变复数时，须先将f或fe变为v再加-es变成复数，即：树叶leafàleaves,半数halfàhalves,妻子wifeàwives,刀knifeàknives,架shelfàshelves,狼wolfàwolves,小偷thiefàthieves.二、直接在词尾加-s的海湾农奴信酋长，悬崖证据上顶房。
【解释】在这些词中，由单数变复数时，直接在词尾加-s构成，即：海湾（gulfàgulfs）农奴（serfàserfs）信（beliefàbeliefs）酋长（chief àchiefs），悬崖（cliffàcliffs）证据（proofàproofs）上顶房（即房顶、屋顶roofàroofs）三、以上两种情况均可的码头侏儒围围巾，手绢复数变二心。
不规则名词复数
fez
fezzes or fezes
fish
fish or fishes
focus
foci or focuses
foot
feet or foot
formula
formulae or formulas
fungus
fungi or funguses
genus
generaorgenuses
mice
louse
Lice(跳蚤)
最后附一张常见的100个不规则复数名词
addendum
addenda or addendums
aircraft
aircraft
alumna
alumnae
alumnus
alumni
analysis
analyses
antenna
antennae or antennas
antithesis
moose
moose
mouse
mice
nebula
nebulae or nebulas
nucleus
nuclei or nucleuses
oasis
oases
offspring
offspring or offsprings
opus
opera or opuses
ovum
ova
ox
oxen or ox
parenthesis
goose
geese
graffito
graffiti
grouse
grouse or grouses
half
halves
hoof
hooves or hoofs
离散数学及其应用重要名词中英对应以及重要概念解释与举例
离散数学及其应用重要名词中英对应以及重要概念解释与举例1 The Foundations: Logic and Proofs（逻辑与证明）1.1 Propositional Logic（命题逻辑）Propositions（命题）——declarative sentence that is either true or false, but not both.判断性语句，正确性唯一。
Truth Table（真值表）Conjunction（合取，“与”，and），Disjunction（析取，or，“相容或”），Exclusive（异或），Negation（非，not），Biconditional（双条件，双向，if and only if）Translating English Sentences1.2 Propositional Equivalences（命题等价）Tautology（永真式、重言式），Contradiction（永假式、矛盾式），Contingency（偶然式）Logical Equivalences（逻辑等价）——Compound propositions that have the same truth values in all possible cases are called logical equivalent.（真值表相同的式子，p<->q是重言式）Logical Equivalences——Page24Disjunctive normal form(DNF，析取范式)Conjunctive normal form(CNF，合取范式) 见Page27～291.3 Predicates and Quantifiers（谓词和量词）Predicates——谓词，说明关系、特征的修饰词Quantifiers——量词Ø Universal Quantifier(全称量词) "全部满足Ø Existential Quantifier(存在量词) $至少有一个Binding Variables(变量绑定，量词作用域与重名的问题)Logical Equivalence Involving QuantifiersNegating Quantified Expressions(量词否定表达：否定全称=存在否定，否定存在=全程否定) Translating from English into Logical Expressions(自然语句转化为逻辑表达)Using Quantifiers in System SpecificationsExamples from Lewis Carrol——全称量词与条件式(p->q)搭配，存在量词与合取式搭配。
爱思唯尔 proof过程
爱思唯尔 proof过程以爱思唯尔PROOF过程为题，我将以人类的视角来叙述这一过程。
在现代科学研究中，PROOF（Patient, Research, Output, Outcome, Follow-up）是爱思唯尔的一种独特方法论，它有助于科学家们精确地进行研究并取得可靠的结果。
下面我将为您详细介绍这一过程。
首先是"P"，也就是病人（Patient）。
在研究开始之前，科学家们首先要确定自己的研究对象，即病人，这可以是人类、动物或细胞等。
病人的选择对于研究的结果至关重要，因此科学家们需要仔细考虑病人的特点、数量和选择方法，以确保研究的可靠性。
接下来是"R"，即研究（Research）。
在进行研究之前，科学家们需要制定研究计划，确定研究的目的、方法和步骤。
他们需要收集和整理相关的文献资料，了解已有的研究成果，并设计实验或观察的具体方案。
科学家们还需要申请研究经费，并进行实验室设备和样本的准备工作。
然后是"O"，即输出（Output）。
在研究过程中，科学家们进行实验、观察和数据分析，他们会收集大量的数据和信息。
通过对数据的处理和分析，科学家们得到了研究的结果和结论。
这些结果可以是新的发现、新的理论或新的技术方法，也可以是对现有知识的补充和修正。
科学家们会将这些研究结果通过学术论文、会议报告、专题讲座等方式进行传播和交流。
接着是"O"，即结果（Outcome）。
研究的结果对于科学家们和学术界来说都非常重要。
科学家们需要对研究结果进行评价和解释，以验证研究的可靠性和有效性。
他们还需要将研究结果与其他研究进行比较和讨论，以进一步推动科学的发展。
此外，研究结果还可以为医学和生物科学提供新的治疗方法和预防策略，从而改善人类的健康和生活质量。
最后是"F"，即随访（Follow-up）。
科学研究并不是一次性的工作，科学家们需要对研究结果进行长期的跟踪和评估。
英语动名形数变化规则
一、名词单数变复数规则1.一般规律：可数名词的复数形式，是在词尾加上“s”。
medals slaves athletes, carnivals awards.注意音变：结尾是清辅音读[s]，结尾是浊辅音或元音读[z]。
2.特殊情况（1）以s、z、x、ch、sh结尾的词，其复数构成在词尾加上“es”。
focuses actresses bushes注意音变：统一加读[iz]。
（2）以“辅音字母+y”结尾的名词，将“y”改变为“i”，再加“es”。
activities discoveries similarities 注意音变：加读[z]【注意区分】journey---journeys play—plays donkey—donkeys（3）①以-o结尾的名词，词后加“es”构成复数。
tomatoes potatoes heroes ②外来词或缩写词，后加“s”构成复数.pianos tobaccos photos kangaroos。
③以“元音+ O”结尾的，词后加“s”构成复数。
注意音变：加读[z]。
（4）以“f”或“fe”结尾的名词，一般把将“f”或“fe”改变为“ves. 如：knives lives leaves但有例外。
roof—roofs belief---beliefs cliff—cliffs gulf---gulfs proof---proofs注意音变：尾音[f]改读[vz]。
3.中学课本中一些特殊的复数形式：如：phenomen---phenomena woman teacher---- women teachers child-- children tooth--- teeth foot--- feet goose--geese mouse---mice ox---oxen basis-- bases analysis--- analyses medium--mdia passers-by lookers-on growns-up stands-by二、动词ing、ed构成规则1.动词后直接加—ing 和-ed.如：earning dreaming performing approaching2.以“e”结尾的单词，去“e”加-ing/ed.如：amusing writing hiding riding3.重读闭音节且有一个辅音字母的，要双写该辅音字母，再加-ing/ed.如：beginning forgetting permitting forbidding fitted quitted fulfilled stopped例外(tie)tying (lie)lying (die)dying (panic) panicking4.辅音字母加y结尾的动词，要变y为i,再加ed.(不能变y为i加ing) 如：reply-replied.三、形容词、副词比较级别构成1.单、双音节的形、副词后直接加“er\ est”构成比较级\ 最高级cooler\ coolest如：deeper\ deepest swifter\ swiftest slighter\ slightest stronger\ strongest simpler\ simplest2.以“e”结尾的分别直接加“r”或“st”构成比较级和最高级如：rarer\ rarest safer\ safest politer\ politest3.重读闭音节，末尾只有一个辅音字母的，双写辅音字母+er--双写辅音字母+est：如：hot\hotter\ hottest big\bigger\ biggest thin\thinner\ thinnest4.以“辅音字母+y”结尾的单、双音节形容词或副词，把“y”变为“i”, 再加“-er\ -est”：如：earlier\ earliest funnier\ funniest happier\ happiest prettier\ prettiest friendlier\ friendliest healthier\ healthiest wealthier\ wealthiest四、形容词+ly构成副词的规则1.在形容词后直接加“ly”如：safely rarely merely fluently excitedly constantly frequently2.“辅音+le”结尾的形容词，去“e”加“y”。
新教材高考英语全程一轮总复习Unit5Poems学生用书新人教版选择性
选择性必修第三册 UNIT 5 POEMS过基础·语基落实(一)分类识记单词——用时少·功效高识记单词写对Ⅰ.知其意1．nursery adj. ____________n. ____________2．nursery rhyme ____________3．diamond n. ____________4．bee n. ____________5．dewdrop n. ____________6．butterfly n. ____________7．lawn n. ____________8．blossom n. ____________9．delicate adj. ____________10．utter vt. ____________adj. ____________11．shelf n. ____________12．cherry n. ____________adj. ____________13．era n. ____________14．prehension n. ____________15．barren adj. ____________16．grief n. ____________17．variation n. ____________18．prejudice n. ____________vt. ____________Ⅱ.写其形1．________ adj.民间的；民俗的；普通百姓的2．________ vt.背诵；吟诵；列举3．________ n．黎明；开端；萌芽4．________ n．业余爱好者adj.业余的；业余爱好的5．________ n. 最后期限；截止日期6．________ n．格式；总体安排；(出版物的)版式vt.格式化7．________ vt.等候；期待；将发生在8．________ adj.空白的；无图画(或韵律、装饰)的；没表情的n．空白；空格9．________ n．版本；(从不同角度的)说法10．________ vt.&vi. 播种；种→____________ (过去式/过去分词/现在分词) 11．________ n．种子；起源；萌芽12．________ conj.在任何地方；在所有……的情况下13．________ adj.复杂的；难懂的14．________ n. 细绳；线；一串vt. 悬挂；系adj. 弦乐器的；线织的→__________ (过去式/过去分词/现在分词)核心单词练透1．They accurately reflect the changes of natural ________(规律)and play an extremely important role in People's Daily life. [2023·威海市模拟] 2．Being the tallest kid in her class, she was ________ (取笑) by her classmates, which made her very annoyed.3．My Chinese teacher offers me some timely and generous guidance on how to ________ (润色)my article.4．At ________ (黎明) they set out and at noon they arrived at their destination.5．As is known to us, body language can give away a lot about your ________ (情绪；心情)．6．In order to improve our spoken English, an English________ (比赛)will be held at 9: 00 am on June 18 in the school meeting room.7．Furthermore, guzheng is of great significance to Chinese ________ (民间的)music leading you to a fantastic world of music.[2021·全国甲卷书面表达] 8．My daughter showed a lot of ________ (同情心) for the people who were begging from the passing shoppers.拓展单词用活【记全记牢】1．drama n．戏；剧；戏剧艺术→________ adj.戏剧(性)的；突然的→________ adv.突然地；戏剧地；引人注目地2．literary adj.文学的；爱好文学的；有文学作品特征的→________ n．文学；文献；著作3．recite vt.背诵；吟诵；列举→________ n. 背诵4．mood n．情绪；心情；语气→________ adj. 喜怒无常的5．respective adj.分别的；各自的→________ adv. 分别；各自；依次为【用准用活】6．All the proof proved his ________，so he was an ________ and straight man.(innocence)7．Judging from her ________ eyes，I knew that her life was full of ________．(sorrow)8．She ________ regularly with her former teacher and their ________ make them more familiar with each other.(correspond)9．I firmly believe that all the ________ are equal and ________ prejudice should be avoided.(race)(二)语境记忆短语——不枯燥·兴趣高记全记牢1.____________ 由……组成(构成)2．____________ 能够控制3．____________ 谋生4．____________ 有意义，讲得通5．____________ 赶上，跟上6．____________ 中肯，扼要；切题7．____________ 试一试8．____________ 除……之外(还有，也)9．____________ 开始，开端；起初10．____________ 成功用准用活选用左栏短语填空1．______________________， swimming is a good form of exercise contributing to our health and strength.2．The world has been developing very fast, and every country must ______________ the time.3．________________ these traditional activities, we have a wider range of choices, such as travelling and visiting our relatives or friends.4．Personally speaking, your ment is very brief and ____________．5．Many people who live along the coast ________________ in fishing industry.·会积累联想拓展·1．以f结尾的名词变复数，直接加s①belief→beliefs 信仰②roof→roofs 屋顶③proof→proofs 证据④chief→chiefs 首领；上司⑤handkerchief→handkerchiefs/handkerchieves 手帕2．“诗歌”家族①poem n. 诗；诗歌②poetry n. 诗歌(总称)③rhyme n. 韵律诗④poet n. 诗人⑤poetic adj. 诗歌的3．如此“成功”①make it 成功②manage to do 设法成功做到③be able to do 做成；做到④succeed in doing 成功做某事⑤be successful in doing sth. 成功做某事(三)仿写用活句式——造佳句·表达高攻重难·解惑释疑练学核心单词1sorrow n．悲伤；悲痛；伤心事vi.感到悲伤(☞sorrowful adj.悲伤的)(1)to one's sorrow 使某人悲伤的是mix joy with sorrow 悲喜交加drown one's sorrow in wine 借酒浇愁(2)sorrow at/over/for sth. 对某事感到悲伤[练透] 单句语法填空/完成句子①He ________ (sorrow) at the news of his friend's accident, his hands shaking. (读后续写情感描写)②As a matter of fact，my nephew daren't look up to see my ________(sorrow) eyes.③一阵悲伤涌上心头，她眼里涌上泪水。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
A SCREENSHOT OF THE OLIVER SYSTEM IN USE .
FIGURE 1
Department of Computer Science, SUNY Stony Brook, St ony Brook, NY 11794-4400, awilden@ Corresponding Author – Pace University, New York City Campus, Computer Science and Information System Department, New York, NY 10028, csharff@
Session S1F OLIVER: AN ONLINE INFERENCE AND VERIFICATION SYSTEM
Andy Wildenberg1 and Christelle Scharff2
Abstract We describe Oliver, the Online inference and verification system for propositional logic. It provides a web-based interface for teaching propositional logic proofs, and accepts any valid direct proof. Oliver provides instant feedback to students as to whether each step is correct or not, encourages experimentation by students and is integrated within the WeBWorK system for online grading and support. Oliver randomizes problems for students to reduce plagiarism and cheating. It has been successfully used to teach propositional logic to several thousand students at SUNY Stony Brook, and is very popular among students and faculty. Index Terms Distance learning, Inference System, Interactive Proofs, Proof Strategy, Proof Verification, Propositional Logic, Web. and students can do the same proof multiple times, experimenting with different strategies of solution to see which are most effective. Problem randomization means that no two students get the same proof, although everyone gets an equivalent proof. This helps students draw the line between helping each other and just copying answers. It has been integrated in the WeBWorK system [14] for online grading and support. Oliver has been used for four semesters at SUNY Stony Brook for teaching propositional logic to more than 2000 computer science, information science and computer engineering students, and will be used at Pace University starting in Fall 2002. Student feedback has been nearly universally positive – students enjoy using the system and feel it teaches them well.
1 2
DESCRIPTION OF THE SYSTEM
Oliver provides a web-based interface for teaching traditional propositional logic. Operators are conjunction (∧), negation (~), inclusive disjunction (∨), implication (→) and double implication ( ). Proofs are of arbitrary complexity, allowing an unlimited number of premises, steps in the proof, and individual statements of any length.
INTRODUCTION
Propositional logic is a core subject in several different fields. According to the ACM standards for computer science education, Discrete Structures is a required course, typically taught in one or two semesters [4,15,16]. One subtopic of Discrete Structures is propositional logic. Proving in propositional logic forms the basis for more complicated proofs in set theory and number theory later on in the Discrete Structures curriculum. Proof-based topics also appear in standard curricula for Mathematics, Philosophy, Electrical Engineering and Computer Engineering. As a new framework for abstract symbol manipulation, propositional logic is typically difficult for students [7-10] – students typically underestimate or overestimate what can be done at any given step within a proof. Cognitive psychologists estimate that only 1 to 4% of the population are able without explicit training to correctly apply the principles of formal logic in situations where it is appropriate to do so [5,12]. Visualization in assisting the reasoning process through the use of software [1,3,11] and web-based can help students learn and enjoy logic [2,3]. Previous interactive systems for doing proofs using strict inference systems and strategies given by the user are Jape [11] and Logic Tutor [1]. We introduce Oliver [13], a web-based system for teaching propositional logic proofs. Students using Oliver enter their proofs online interactively one step at a time with the system checking each step. Feedback is instantaneous,
0-7803-7444-4/02/$17.00 © 2002 IEEE November 6 - 9, 2002, Boston, MA 32 nd ASEE/IEEE Frontiers in Education Conference S1F-22
Session S1F
Oliver is a unique system in several ways. Proofs are done with standard operators compatible with the systems taught by most discrete mathematics textbooks. At SUNY Stony Brook, Oliver was used coupled with the theory taught in [6]. Currently, the help system for Oliver is a single web page that introduces the symbols for the different logical operators and shows a sam