E.Triesch, A note on law of large numbers for fuzzy variables

湖北省随州市行知高级中学2023-2024学年高一下学期3月月考英语试题一、听力选择题1．How did the speakers go to the capital?A．By train.B．By bus.C．By car.2．What is the girl going to do?A．Go shopping.B．Go swimming.C．Go camping.3．Why is Jenny unable to go out with the man?A．She is walking the dog.B．She is looking after her cousins.C．She needs to do her homework.4．What does the woman want to have tonight?A．Beef and chicken.B．Fruit and vegetables.C．Fish and beef. 5．Where does the conversation probably take place?A．In the car.B．At school.C．In the supermarket.听下面一段较长对话，回答以下小题。

6．What would the man like to know about the future?A．How the environment will be.B．How fast the technology will develop.C．How rich he will be.7．What might the woman want to be if she can choose again?A．A scientist.B．A teacher.C．A doctor.听下面一段较长对话，回答以下小题。

8．How many rooms will Jessica book?A．Three.B．Six.C．Eight.9．What day is it today?A．Wednesday.B．Thursday.C．Friday.10．Who will pick up the visitors?A．Evan.B．Jessica.C．Henry.听下面一段较长对话，回答以下小题。

剑桥雅思阅读10真题解析(test2)雅思阅读部分的真题资料，同学们需要进行一些细致的总结，比如说解析其实就是很重要的内容，接下来就是店铺给同学们带来的关于剑桥雅思阅读10真题解析(test2)的内容，一起来详细的分析一下吧，希望对你们的备考有所帮助。

剑桥雅思阅读10原文(test2)READING PASSAGE 1You should spend about 20 minutes on Questions 1-13, which are based on Reading Passage 1 on the following pages.Questions 1-7Reading Passage 1 has seven paragraphs, A-G.Choose the correct heading for each paragraph from the list of headings below.Write the correct number，i-ix，in boxes 1-7 on your answer sheetList of Headingsi The search for the reasons for an increase in populationii Industrialisation and the fear of unemploymentiii The development of cities in Japaniv The time and place of the Industrial Revolutionv The cases of Holland, France and Chinavi Changes in drinking habits in Britainvii Two keys to Britain’s industrial revolutionviii Conditions required for industrialisationix Comparisons with Japan lead to the answer1 Paragraph A2 Paragraph B3 Paragraph C4 Paragraph D5 Paragraph E6 Paragraph F7 Paragraph GTea and the Industrial RevolutionA Cambridge professor says that a change in drinking babits was the reason for the Industrial Revolution in Britain. Anjana Abuja reportsA Alan Macfarlane, professor of anthropological science at King’s College, Cambridge, has, like other historians, spent decades wrestling with the enigma of the Industrial Revolution. Why did this particular Big Bang — the world-changing birth of industry — happen in Britain? And why did it strike at the end of the 18th century?B Macfarlane compares the puzzle to a combination lock. ‘There are about 20 different factors and all of them need to be present before the revolution can happen,’ he says. For industry to take off, there needs to be the technology and power to drive factories, large urban populations to provide cheap labour, easy transport to move goods around, an affluent middle-class willing to buy mass-produced objects, a market-driven economy and a political system that allows this to happen. While this was the case for England, other nations, such as Japan, the Netherlands and France also met some of these criteria but were not industrialising. ‘All these factors must have been necessary but not sufficient to c ause the revolution,’ says Macfarlane. ‘After all, Holland had everything except coal while China also had many of these factors. Most historians are convinced there are one or two missing factors that you need to open the lock.’C The missing factors, he proposes, are to be found in almost even kitchen curpboard. Tea and beer, two of the nation’sfavourite drinks, fuelled the revolution. The antiseptic properties of tannin, the active ingredient in tea, and of hops in beer — plus the fact that both are made with boiled water — allowed urban communities to flourish at close quarters without succumbing to water-borne diseases such as dysentery. The theory sounds eccentric but once he starts to explain the detective work that went into his deduction, the scepticism gives way to wary admiration. Macfarlane’s case has been strengthened by support from notable quarters — Roy Porter, the distinguished medical historian, recently wrote a favourable appraisal of his research.D Macfarlane had wondered for a long time how the Industrial Revolution came about. Historians had alighted on one interesting factor around the mid-18th century that required explanation. Between about 1650 and 1740，the population in Britain was static. But then there was a burst in population growth. Macfarlane says: ‘The infant mortality rate halved in the space of 20 years, and this happened in both rural areas and cities, and across all classes. People suggested four possible causes. Was there a sudden change in the viruses and bacteria around? Unlikely. Was there a revolution in medical science? But this was a century before Lister’s revolution_ Was there a change in environmental conditions? There were improvements in agriculture that wiped out malaria, but these were small gains. Sanitation did not become widespread until the 19th century. The only option left is food. But the height and weight statistics show a decline. So the food must have got worse. Efforts to explain this sudden reduction in child deaths appeared to draw a blank.’E This population burst seemed to happen at just the right time to provide labour for the Industrial Revolution. ‘When youstart moving towards an industrial revolution, it is economically efficient to have people living close together,’ says Macfarlane. ‘But then you get disease, particularly from human waste.’ Some digging around in historical records revealed that there was a change in the incidence of water-borne disease at that time, especially dysentery. Macfarlane deduced that whatever the British were drinking must have been important in regulating disease. He says, ‘We drank beer. For a long time, the English were protected by the strong antibacterial agent in hops, which were added to help preserve the beer. But in the late 17th century a tax was introduced on malt, the basic ingredient of beer. The poor turned to water and gin and in the 1720s the mortality rate began to rise again. Then it suddenly dropped again. What caused this?’F Macfarlane looked to Japan, which was also developing large cities about the same time, and also had no sanitation. Water-borne diseases had a much looser grip on the Japanese population than those in Britain. Could it be the prevalence of tea in their culture? Macfarlane then noted that the history of tea in Britain provided an extraordinary coincidence of dates. Tea was relatively expensive until Britain started a direct clipper trade with China in the early 18th century. By the 1740s, about the time that infant mortality was dipping, the drink was common. Macfarlane guessed that the fact that water had to be boiled, together with the stomach-purifying properties of tea meant that the breast milk provided by mothers was healthier than it had ever been. No other European nation sipped tea like the British, which, by Macfarla ne’s logic, pushed these other countries out of contention for the revolution.G But, if tea is a factor in the combination lock, why didn’tJapan forge ahead in a tea-soaked industrial revolution of its own? Macfarlane notes that even though 17th-century Japan had large cities, high literacy rates, even a futures market, it had turned its back on the essence of any work-based revolution by giving up labour-saving devices such as animals, afraid that they would put people out of work. So, the nation that we now think of as one of the most technologically advanced entered the 19th century having ‘abandoned the wheel’._oseph Lister was the first doctor to use antiseptic techniques during surgical operations to prevent infections.Questions 8-13Do the following statements agree with the information given in Reading Passage 1?In boxes 8-13 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this8 China’s transport system was not suitable for industry in the 18th century.9 Tea and beer both helped to prevent dysentery in Britain.10 Roy Porter disagrees with Professor Macfarlane’s findings.11 After 1740，there was a reduction in population in Britain.12 People in Britain used to make beer at home.13 The tax on malt indirectly caused a rise in the death rate.READING PASSAGE 2You should spend about 20 minutes on Questions 14-26, which are based on Reading Passage 2 below.Gifted children and learningA Internationally, ‘giftedness’ is most frequentlydetermined by a score on a general intelligence test, known as an IQ test, which is above a chosen cutoff point, usually at around the top 2-5%. Children’s educational environment contributes to the IQ score and the way intelligence is used. For example, a very close positive relationship was found when children’s IQ scores were compared with their home educational provision (Freeman, 2010). The higher the children’s IQ scores, especially over IQ 130, the better the quality of their educational backup, measured in terms of reported verbal interactions with parents, number of books and activities in their home etc. Because IQ tests are decidedly influenced by what the child has learned, they are to some extent measures of current achievement based on age-norms; that is, how well the children have learned to manipulate their knowledge and know-how within the terms of the test. The vocabulary aspect, for example, is dependent on having heard those words. But IQ tests can neither identify the processes of learning and thinking nor predict creativity.B Excellence does not emerge without appropriate help. To reach an exceptionally high standard in any area very able children need the means to learn, which includes material to work with and focused challenging tuition — and the encouragement to follow their dream. There appears to be a qualitative difference in the way the intellectually highly able think, compared with more average-ability or older pupils, for whom external regulation by the teacher often compensates for lack of internal regulation. To be at their most effective in their self-regulation, all children can be helped to identify their own ways of learning —metacognition —which will include strategies of planning, monitoring, evaluation, and choice of what to learn. Emotional awareness is also part of metacognition, so children should behelped to be aware of their feelings around the area to be learned, feelings of curiosity or confidence, for example.C High achievers have been found to use self-regulatory learning strategies more often and more effectively than lower achievers, and are better able to transfer these strategies to deal with unfamiliar tasks. This happens to such a high degree in some children that they appear to be demonstrating talent in particular areas. Overviewing research on the thinking process of highly able children, (Shore and Kanevsky, 1993) put the instructor’s problem succinctly: ‘If they [the gifted] merely think mo re quickly, then we need only teach more quickly. If they merely make fewer errors, then we can shorten the practice’. But of course, this is not entirely the case; adjustments have to be made in methods of learning and teaching, to take account of the many ways individuals think.D Yet in order to learn by themselves, the gifted do need some support from their teachers. Conversely, teachers who have a tendency to ‘overdirect’ can diminish their gifted pupils’ learning autonomy. Although ‘spoon-feeding’ can produce extremely high examination results, these are not always followed by equally impressive life successes. Too much dependence on the teachers risks loss of autonomy and motivation to discover. However, when teachers help pupils to reflect on their own learning and thinking activities, they increase their pupils’ self-regulation. For a young child, it may be just the simple question ‘What have you learned today?’ which helps them to recognise what they are doing. Given that a fundamental goal of education is to transfer the control of learning from teachers to pupils, improving pupils’ learning to learn techniques should be a major outcome of the school experience,especially for the highly competent. There are quite a number of new methods which can help, such as child-initiated learning, ability-peer tutoring, etc. Such practices have been found to be particularly useful for bright children from deprived areas.E But scientific progress is not all theoretical, knowledge is a so vital to outstanding performance: individuals who know a great deal about a specific domain will achieve at a higher level than those who do not (Elshout, 1995). Research with creative scientists by Simonton (1988) brought him to the conclusion that above a certain high level, characteristics such as independence seemed to contribute more to reaching the highest levels of expertise than intellectual skills, due to the great demands of effort and time needed for learning and practice. Creativity in all forms can be seen as expertise mixed with a high level of motivation (Weisberg, 1993).F To sum up, learning is affected by emotions of both the individual and significant others. Positive emotions facilitate the creative aspects of learning and negative emotions inhibit it. Fear, for example, can limit the development of curiosity, which is a strong force in scientific advance, because it motivates problem-solving behaviour. In Boekaerts’ (1991) review of emotion the learning of very high IQ and highly achieving children, she found emotional forces in harness. They were not only curious, but often had a strong desire to control their environment, improve their learning efficiency and increase their own learning resources.Questions 14-17Reading Passage 2 has six paragraphs, A-F.Which paragraph contains the following information?Write the correct letter, A-F, in boxes 14-17 on your answersheet.NB You may use any letter more than once.14 a reference to the influence of the domestic background on the gifted child15 reference to what can be lost if learners are given too much guidance16 a reference to the damaging effects of anxiety17 examples of classroom techniques which favour socially-disadvantaged childrenQuestions 18-22Look at the following statements (Questions 18-22) and the list of people below.Match each statement with the correct person or people, A-E.Write the correct letter, A-E, in boxes 18-22 on your answer sheet.18 Less time can be spent on exercises with gifted pupils who produce accurate work.19 Self-reliance is a valuable tool that helps gifted students reach their goals.20 Gifted children know how to channel their feelings to assist their learning.21 The very gifted child benefits from appropriate support from close relatives.22 Really successful students have learnt a considerable amount about their subject.List of PeopleA FreemanB Shore and KanevskyC ElshoutD SimontonE BoekaertsQuestions 23-26Complete the sentences below.Choose NO MORE THAN TWO WORDS from the passage for each answer.Write your answers in boxes 23-26 on your answer sheet23 One study found a strong connection between children’s IQ and the availability of andat home.24 Children of average ability seem to need more direction from teachers because they do not have25 Metacognition involves children understanding their own learning strategies, as well as developing26 Teachers who rely on what is known as often produce sets of impressive grades in class tests.READING PASSAGE 3You should spend about 20 minutes on Questions 27-40, which are based on Reading Passage 3 below.Museums of fine art and their publicThe fact that people go to the Louvre museum in Paris to see the original painting Mona Lisa when they can see a reproduction anywhere leads us to question some assumptions about the role of museums of fine art in today’s worldOne of the most famous works of art in the world is Leonardo da Vinci’s Mona Lisa. Nearly everyone who goes to see the original will already be familiar with it from reproductions, but they accept that fine art is more rewardingly viewed in its original form.However, if Mona Lisa was a famous novel, few people wouldbother to go to a museum to read the writer’s actual manuscript rather than a printed reproduction. This might be explained by the fact that the novel has evolved precisely because of technological developments that made it possible to print out huge numbers of texts, whereas oil paintings have always been produced as unique objects. In addition, it could be argued that the practice of interpre ting or ‘reading’ each medium follows different conventions. With novels, the reader attends mainly to the meaning of words rather than the way they are printed on the page, whereas the ‘reader’ of a painting must attend just as closely to the material form of marks and shapes in the picture as to any ideas they may signify.Yet it has always been possible to make very accurate facsimiles of pretty well any fine art work. The seven surviving versions of Mona Lisa bear witness to the fact that in the 16th century, artists seemed perfectly content to assign the reproduction of their creations to their workshop apprentices as regular ‘bread and butter’ work. And today the task of reproducing pictures is incomparably more simple and reliable, with reprographic techniques that allow the production of high-quality prints made exactly to the original scale, with faithful colour values, and even with duplication of the surface relief of the painting.But despite an implicit recognition that the spread of good reproductions can be culturally valuable, museums continue to promote the special status of original work.Unfortunately, this seems to place severe limitations on the kind of experience offered to visitors.One limitation is related to the way the museum presents its exhibits. As repositories of unique historical objects, art museumsare often called ‘treasure houses’. We are reminded of this even before we view a collection by the presence of security guards, attendants, ropes and display cases to keep us away from the exhibits. In many cases, the architectural style of the building further reinforces that notion. In addition, a major collection like that of London’s National Gallery is housed in numerous rooms, each with dozens of works, any one of which is likely to be worth more than all the average visitor possesses. In a society that judges the personal status of the individual so much by their material worth, it is therefore difficult not to be impressed by one’s own relative ‘worthlessness’ in such a n environment.Furthermore, consideration of the ‘value’ of the original work in its treasure house setting impresses upon the viewer that, since these works were originally produced, they have been assigned a huge monetary value by some person or institution more powerful than themselves. Evidently, nothing the viewer thinks about the work is going to alter that value, and so today’s viewer is deterred from trying to extend that spontaneous, immediate, self-reliant kind of reading which would originally have met the work.The visitor may then be struck by the strangeness of seeing such diverse paintings, drawings and sculptures brought together in an environment for which they were not originally created. This ‘displacement effect’ is further heightened by the sheer volume of exhibits. In the case of a major collection, there are probably more works on display than we could realistically view in weeks or even months.This is particularly distressing because time seems to be a vital factor in the appreciation of all art forms. A fundamental difference between paintings and other art forms is that there isno prescribed time over which a painting is viewed. By contrast, the audience encounters an opera or a play over a specific time, which is the duration of the performance. Similarly, novels and poems are read in a prescribed temporal sequence, whereas a picture has no clear place at which to start viewing, or at which to finish. Thus art works themselves encourage us to view them superficially, without appreciating the richness of detail and labour that is involved.Consequently, the dominant critical approach becomes that of the art historian, a specialised academic approach devoted to ‘discovering the meaning’ of art within the cultural context of its time. T his is in perfect harmony with the museum’s function, since the approach is dedicated to seeking out and conserving ‘authentic’, ‘original’ readings of the exhibits. Again, this seems to put paid to that spontaneous, participatory criticism which can be found in abundance in criticism of classic works of literature, but is absent from most art history.The displays of art museums serve as a warning of what critical practices can emerge when spontaneous criticism is suppressed. The museum public, like any other audience, experience art more rewardingly when given the confidence to express their views. If appropriate works of fine art could be rendered permanently accessible to the public by means of high-fidelity reproductions, as literature and music already are, the public may feel somewhat less in awe of them. Unfortunately, that may be too much to ask from those who seek to maintain and control the art establishment.Questions 27-31Complete the summary using the list of words, A-L, below.Write the correct letter, A-L, in boxes 27-31 on your answersheet.The value attached to original works of artPeople go to art museums because they accept the value of seeing an original work of art. But they do not go to museums to read original manuscripts of novels, perhaps because the availability of novels has depended on 27 for so long, and also because with novels, the 28 are the most important thing.However, in historical times artists such as Leonardo were happy to instruct 29 to produce copies of their work and these days new methods of reproduction allow excellent replication of surface relief features as well as colour and 30It is regrettable that museums still promote the superiority of original works of art, since this may not be in the interests of the 31A institutionB mass productionC mechanical processesD publicE paintsF artistG size H underlying ideas I basic technologyJ readers K picture frames L assistantsQuestions 32-35Choose the correct letter, A, B, C or D.Write the correct letter in boxes 32-35 on your answer sheet32 The writer mentions London’s National Gallery to illustrateA the undesirable cost to a nation of maintaining a huge collection of art.B the conflict that may arise in society between financial and artistic values.C the n egative effect a museum can have on visitors’ opinions of themselves.D the need to put individual well-being above large-scaleartistic schemes.33 The writer says that today, viewers may be unwilling to criticise a work becauseA they lack the knowledge needed to support an opinion.B they fear it may have financial implications.C they have no real concept of the work’s value.D they feel their personal reaction is of no significance.34 According to the writer, the ‘displacement effect’ on the visitor is caused byA the variety of works on display and the way they are arranged.B the impossibility of viewing particular works of art over a long period.C the similar nature of the paintings and the lack of great works.D the inappropriate nature of the individual works selected for exhibition.35 The writer says that unlike other forms of art, a painting does notA involve direct contact with an audience.B require a specific location for a performance.C need the involvement of other professionals.D have a specific beginning or end.Questions 36-42Do the following statements agree with the views of the writer in Reading Passage 3?In boxes 36-40 on your answer sheet, writeYES if the statement agrees with the views of the writerNO if the statement contradicts the views of the writerNOT GIVEN if it is impossible to say what the writer thinksabout this36 Art history should focus on discovering the meaning of art using a range of media.37 The approach of art historians conflicts with that of art museums.38 People should be encouraged to give their opinions openly on works of art.39 Reproductions of fine art should only be sold to the public if they are of high quality.40 In the future, those with power are likely to encourage more people to enjoy art.剑桥雅思阅读10原文参考译文(test2)Passage 1参考译文：茶与工业革命一个剑桥教授称英国工业革命的导火索是饮水习性的改变。

小学上册英语基本全练全测(有答案)英语试题一、综合题(本题有100小题，每小题1分，共100分.每小题不选、错误，均不给分)1.The birds are ___ in the trees. (chirping)2.The ____ is a tiny insect that helps pollinate flowers.3.My favorite fruit is ________ (苹果).4.Solutions can be classified as acidic, basic, or ______.5.The first person to fly solo nonstop across the Atlantic was _______ Lindbergh.6.I can ___ (draw) a cat.7.They are going to ________ a movie.8.World War II ended in _______. (1945年)9.My pet fish swims in _______ (圆形) patterns.10.My pet rabbit has big _______ (耳朵).11.The ______ (小鹿) grazes quietly in the ______ (草地).12.I enjoy hiking to see beautiful ______ (风景) and wildlife.13.Ionic compounds are formed from ______ and nonmetals.14. A ________ is an area of land that is controlled by a government.15.We are going to ______ (camp) in the woods.16. A ________ is small and furry.17.The ________ (草原) is home to many wild animals.18.The __________ (历史的深刻理解) informs perspectives.19. A ________ (植物实验室) conducts research.20.What do we call the process of making a plan to achieve a goal?A. Goal-settingB. PlanningC. StrategizingD. All of the above答案: D. All of the above21.What do you call the middle of the Earth?A. CrustB. MantleC. CoreD. Shell答案: C22.The _______ (鳄鱼) lives in the swamp.23.My friend’s family is very ____.24.The _______ of a substance is a measure of how much mass is contained in a given volume.25.The sun shines _____ (brightly/dimly).26.I have a toy ________ that can fly.27.green initiatives) promote sustainability in cities. The ____28.What is the smallest unit of life?A. CellB. TissueC. OrganD. Organism答案: A. Cell29.ts are __________ (有毒的) and should be handled carefully. Some pla30.Animals that live in the Arctic are adapted to ______ weather.31.The dog loves to chase ______ (balls).32.The chemical formula for ammonia is __________.33.What is the capital of Italy?A. RomeB. FlorenceC. NaplesD. Venice答案: A. Rome34. (Revolutionary) War was fought for American independence. The ____35. A ______ is a substance that can conduct electricity.36.The lemonade is ___. (cold)37.The chemical process that produces energy in our cells is called ________.38.My ______ travels around the world for work.39. A _____ (植物保护措施) can help preserve endangered species.40.The chemical symbol for barium is _______.41.I enjoy _____ (弹奏) the piano.42.Which of these is a popular fruit?A. SpinachB. CarrotC. StrawberryD. Broccoli答案:C43.The _____ (sky) is blue.44.The weather is _____ outside today. (nice)45.The invention of ________ changed transportation forever.46.The process where a solid turns directly into gas is called ______.47.The Earth's surface is home to diverse forms of ______.48.My brother is a ______. He enjoys coding games.49.The Earth's axis is tilted at an angle of ______ degrees.50.What is the name of the famous art museum in Paris?A. LouvreB. British MuseumC. Uffizi GalleryD. Rijksmuseum 答案:A51.I want to ________ (visit) the zoo.52.The ancient Romans used ________ for transportation of goods.53. A ________ (生态平衡) ensures diversity.54.I have a toy ________ (玩具名称) that can stack high.55. A river flows into a __________.56.中国的________ (philosophy) 包括儒家、道家等不同流派。

专题六2020浙江卷1月-2020年高考阅读理解真题词汇和长难句解读（解析版）2020浙江卷1月阅读理解AI never knew anyone who’d grown up in Jackson without being afraid of Mrs. Calloway, our librarian. She ran Jackson’s Carnegie Library absolutely by herself. SILENCE in big black letters was on signs hung everywhere. If she thought you were dressed improperly, she sent you straight back home to change your clothes. I was willing; I would do anything to read.My mother was not afraid of Mrs. Calloway. She wished me to have my own library card to check out books for myself. She took me in to introduce m e. “Eudora is nine years old and has my permission to read any book she wants from the shelves, children or adults,” Mother said.Mrs. Calloway made her own rules about books. You could not take back a book to the library on the same day you’d taken it out; it made no difference to her that you’d read every word in it and needed another to start. You could take out two books at a time and two only. So two by two, I read library books as fast as I could go, rushing them home in the basket of my bicycle. From the minute I reached our house, I started to read. I knew this was extreme happiness, knew it at the time.My mother shared this feeling of mine. Now, I think of her as reading so much of the time while doing something else. I remember her reading a magazine while taking the part of the Wolf in a game of “Little Red Riding H ood” with my brother’s two daughters. She’d just look up at the right time, long enough to answer —in character —“The better to eat you with, my dear,” and go back to her place in the magazine article.21. Which of the following best describes Mrs. Calloway?A. Quiet.B. Strict.C. Humorous.D. Considerate.22. What do the underlined words “this feeling” refer to in the last paragraph?A. Desire to read.B. Love for Mrs. Calloway.C. Interest in games.D. Fear of the library rules.23. Where is the text probably from?A. A guidebook.B. An autobiography.C. A news report.D. A book review.【答案】21.B 22.A 23.B【词汇】1. absolutely2. improperly3. permission4. extreme5. humorous6. considerate7. desire 8. autobiography1. absolutely adv. 完全地；绝对地；[语]独立地int. （表示赞同）完全正确，一点没错2. improperly adv. 不正确地；不适当地3. permission n. 许可；允许；同意4. extreme adj. 极度的；极端的n. 极端；极限5. humorous adj. 幽默的；诙谐的6. considerate adj. 考虑周到的；体谅的；体贴的7. desire n. 渴望；愿望；欲望v. 渴望；向往；要求8. autobiography n. 自传【长难句】1. Eudora is nine years old and has my permission to read any book she wants from the shelves, children or adults. 【句子分析】主干部分：Eudora is nine years old and has my permission to read any book定语从句：she wants from the shelves,省略that 修饰book【翻译】尤多拉九岁了，我允许她从书架上读任何一本书，无论是儿童还是成人。

8Laws of large numbers8.1IntroductionThe following comes up very often,especially in statistics.We have an exper-iment and a random variable X associated with it.We repeat the experiment n times and let X1,X2,···,X n be the values of the random variable we get. We can think of the n-fold repetition of the original experiment as a sort of super-experiment and X1,X2,···,X n as random variables on for it.We assume that the experiment does not change as we repeat it,so that the X j are indentically distributed.(Recall that this means that X i and X j have the same pmf or pdf.)And we assume that the different perfomances of the experiment do not influence each other.So the random variables X1,···,X n are independent.In statstics one typically does not know the pmf or the pdf of the X j. The statistician’s job is to take the random sample X1,···,X n and make conclusions about the distribution of X.For example,one would like to know the mean,E[X],of X.The simplest way to estimate it is to look at the“sample mean”which is defined to beX n=1nn i X iNote that X n is itself a random variable.Intuitively we expect that as n→∞,X n will converge to E[X].What exactly do we mean by“convergence”of a sequence of random variables?And what can we say about the rate of convergence and the error?These questions are the focus of this chapter.We already know quite a bit about X n.Its mean isµ=E[X].And its variance isσ2/n whereσ2is the common variance of the X j.Thefirst theorems in this chapter will say that as n→∞,X n converges to the constantµin some sense.Results like this are called a“law of large numbers.”We will see two of them corresponding to two different notions of convergence. The other big theorem of this chapter is the central limit theorem.Suppose we shift and rescale X n byX n−µσ/√nNote that the above random variable has mean zero and variance one.The1CLT says that it converges to a standard normal under some very mild as-sumptions on the distribution of X.8.2Weak law of large numbersIf we roll a fair six-sided die,the mean of the number we get is3.5.If we roll the die a large number of times and average the numbers we get(i.e.,compute X n),then we do not expect to get exactly3.5,but rather something close.So we could ask if|X n−3.5|<0.01.This is an event(for the super-experiment), so we can consider its probability:|X n−3.5|<0.01In particular we might expect this probability to go to zero as n→∞.This motivates the following definition.Definition1.Let Y n be a sequence of random variables,and Y a random variable,all defined on the same probability space.We say Y n converges to Y in probability if for everyǫ>0,limn→∞P(|Y n−Y|>ǫ)=0Theorem1.(Weak law of large numbers)Let X j be an i.i.d.sequence withfinite mean and variance.Letµ=E[X j].ThenX n=1nnj=1X j→µin probabilityThere are better versions of the theorem in the sense that they have weaker hypotheses(you don’t need to assume the variance isfinite).There is also a stronger theorem that has a stronger form of convergence(strong law of large numbers).We will eventually prove the theorem,butfirst we introduce another notion of convergence.Definition2.Let Y n be a sequence of random variables withfinite variance and Y a random variable withfinite variance.We say that Y n converges to Y in mean square iflimn→∞E[(Y n−Y)2]=02In analysis this is often called convergence in L2.Proposition1.Let X n be a sequence of i.i.d.random variables withfinite variance.Letµ=E[X n].Then X n coverges toµin mean square.Proof.We have to showlimn→∞E[(X n−µ)2]=0But since the mean of X n isµ,E[(X n−µ)2]is the variance of X n.We know that this variance isσ2/n which obviously goes to zero as n→∞.Next we show that convergence in mean square implies convergence in probability.The tool to show this is the following inequality:Proposition2.(Chebyshev’s inequality)P(|X|≥a)≤E[X2] a2Proof.To make things concrete we assume we have a continuous RV.Then letting f(x)be the pdf of X,E[X2]= ∞−∞x2f(x)dxSince the integrand is non-negative,E[X2]≥ |x|≥a x2f(x)dx≥ |x|≥a a2f(x)dx=a2 |x|≥a f(x)dx=a2P(|X|≥a)Thus we have the inequality in the proposition.Proposition3.Let Y n is a sequence of random variables withfinite variance and Y is a random variable withfinite variance.Suppose Y n converges to Y in mean square.Then it converges in probability.3Proof.Let ǫ>0.We must showlim n →∞P (|Y n −Y |>ǫ)=0By Chebyshev’s inequality,P (|Y n −Y |>ǫ)≤E [(Y n −Y )2]ǫ2By hypothesis E [(Y n −Y )2]→0as n →∞.So for a fixed ǫ,E [(Y n −Y )2]/ǫ2→0as n →∞.8.3Central limit theoremLet X n be an i.i.d.sequence with finite variance.Let µthe their common mean and σ2their common variance.As before we let X n =1n n j =1X j .Define Z n =X n −µσ/√n = n j =1X j −nµ√nσNote that E [Z n ]=0,var (Z n )=1.The best way to remember the definition is that it is X n shifted and scaled so that it has mean 0and variance 1.The central limit theorem says that the distribution of Z n converges to a standard normal.There are several senses in which it might converge,so we have to make this statement more precise.We might ask if the density function of Z n converges to that of a standard normal,ie.,1√2πexp(−z 2/2).We do not assume that X n is a continuous RV.If it is a discrete RV,then so is Z n .So it does not even have a density function.Let Φ(x )be the cdf of a standard normal:Φ(x )= x −∞1√2πe −z 2/2dz Theorem 2.(Central limit theorem )Let X n be an i.i.d.sequence of random variables with finite mean µand variance σ2.Define Z n as above.Then for all x ∈Rlim n →∞P (Z n ≤x )=Φ(x )4In words the theorem says that the cdf of Z n converges pointwise to the cdf of the standard normal.This is an example of what is called “convergence in distribution”in probability.However,we caution the reader that the general definitionof convergence in distbribution involves some technicalities.This is the most important theorem in the course and plays a major role in statistics.In particular much of the theory of confidence intervals and hypothesis testing is the central limit theorem in disguise.Example:A computer has a random number generator that generates ran-dom numbers uniformly distributed in [0,1].We run it 100times and let S n be the sum of the 100numbers.Estimate P (S n ≥52).Let X j be the numbers.So E [X j ]=1/2and var (X j )=1/12.SoZ n =S n −nµ√nσ=S n −100×1/210 1/12So S n ≥52is the same asZ n ≥52−100×1/210 1/12=2√1210=0.6928So P (S n ≥52)=P (Z n ≥0.6928).Now we approximate P (Z n ≥0.6928)by assuming Z n has a standard normal distribution.Then we can “look up”P (Z n ≥0.6928)in R.We use P (Z n ≥0.6928)=1−P (Z n ≤0.6928).In R we get P (Z n ≤0.6928)from pnorm (0.6928).(This is the same as pnorm (0.6928,0,1).If you don’t specify a mean and variance when you call pnorm ()it assumes a standard normal.)So we findP (S n ≥52)=P (Z n ≥0.6928)=1−P (Z n ≤0.6928)≈1−0.7548=0.2442Example:We flip a fair coin n times.How large must n be to have P (|(fraction of H )−1/2|≥0.01)≤0.05.According to R,qnorm (0.975)=1.959964.This means that for a stan-dard normal Z P (Z ≤1.959964)=0.975.By symmetry,P (|Z |≤1.959964)=0.95.So P (|Z |≥1.959964)=0.05.Let X j be 1if there is H on j th flip,0if it is T.Let X n be the usual.Then X n is the fraction of heads.For a single flip the variance is p (1−p )=1/4.So σ=1/2.SoZ n =X n −1/2σ√n =X n −1/2√n/25So|X n−1/2|≥0.01is the same as|Z n|≥0.01nσ√n,i.e.,|Z n|≥0.02√n.So P(|(fraction of H)−1/2|≥0.01)≤0.05=P(|Z n|≥0.02√n)So it this is to be less than0.05we need0.02√n≥2.So n≥10,000.End of Nov9lectureConfidence intervals:The following is an important problem in statistics. We have a random variable X(usually called the population).We know its varianceσ2,but we don’t know its meanµ.We have a“random sample,”i.e.,random variables X1,X2,···,X n which are independent random varibles which all have the sample distribution as X.We want to use our one sample X1,···,X n to estimateµ.The natural estimate forµis the sample meanX n=1nn j=1X jHow close is X n to the true value ofµ?This is a vague question.We make it precise as follows.For whatǫ>0will P(|X n−µ|≤ǫ)=0.95?We say the[X n−ǫ,X n+ǫ]is a95%confidence interval forµ.(The choice of95% is somewhat arbitrary.We can use98%for example.If n is large we can use the CLT tofigure out whatǫshould be.As before we letZ n=X n−µσ/√nSo|X n−µ|≤ǫis equivalent to|Z n|≤ǫ√n/σSo we wantP(|Z n|≤ǫ√n/σ)=0.95The CLT says that the distribution for Z n is approximately that of a standard normal.If Z is a standard normal,then P(|Z|≤1.96)=0.95.Soǫ√n/σ= 1.96.So we have found that the95%confidence interval forµis[µ−ǫ,µ+ǫ] whereǫ=1.96∗σ/√nExample:Lightbulbs6The 1/2game :Suppose X 1,X 2,···are i.i.d.and integer valued.For example,we flip a coin repeatedly and let X n be 1if we get heads on the n th flip and 0if we get tails.We want to estimate P (a ≤S n ≤b ),where S n = n j =1X j .We take a and b to be integers.ThenP (a −δ≤S n ≤b +δ)=P (a ≤S n ≤b )for any positive δ<1.But the CLT will give us different answers depending on which δwe use.What is the best?Answer δ=1/2.So we approximate P (a ≤S n ≤b )byP (a −1/2≤S n ≤b +1/2)=P (a −1/2−nµσ√n ≤S n −nµσ√n ≤b +1/2−nµσ√n)≈P (a −1/2−nµσ√n ≤Z ≤b +1/2−nµσ√n)where Z is a standard normal.End of Nov 18lectureExample:Let X n be an i.i.d.sequence of standard normals.LetY n =1n nj =1X 2jUse the CLT to find n so that P (Y n ≥1.01)=5%.Note that the the i.i.d.sequence that we apply the CLT to is X 2n not X n .The mean of X 2n is E [X 2n ]=1.The variance is E [X 4n ]−(E [X 2n ])2=3−1=2.The mean of Y n is 1n n =1.The variance isvar (Y n )=1n 2n 2So to standardize,Z n =Y n −1 2/n7SoP(Y n≥1.01)=P(Y n−12/n≥1.01−1 2/n)≈P(Z≥1.01−12/n)where Z is a standard normal.R says that qnorm(0.95)=1.645.So P(Z≥1.645)=0.05.So0.012/n=1.645,n=(1.645)2,000=54,1218.4Strong law of large numbersSuppose X n is an i.i.d.sequence.As beforeX n=1nn j=1X jThe weak law involves probabilities that X n does certain things.We now ask if lim n→∞X n exists and what does it converge to.Each X n is a random variable,i.e.,a function ofω,a point in the sample space.It is possible that this limit converges toµfor someωbut not for otherω.Example Weflips a fair coin infinitely often.Let X n be1if the n thflip is heads,0if it is tails.Then X n is the fraction of heads in thefirst n flips.We can think of the sample space as sequences of H’s and T’s.So one possibleωis the sequence with all H’s.For thisω,X n(ω)=1for all n.So X n(ω)=1for all n.So lim n→∞X n exists,but it equals1,not the mean of X n which is1/2.So it is certainly not true that lim n→∞X n=µfor allω. Our counterexample where we get all heads is“atypical.”So we might hope that the set ofωfor which the limit of the X n is notµis“small”.Definition3.Let Y n be a sequence of random variables and Y a random variable.We say X n converges to X with probability one ifP({ω:limn→∞Y n(ω)=Y(ω)})=1More succintly,P(Y n→Y)=18This is a stronger form of convergence than convergence in probability. (This is not at all obvious.)Theorem3.If Y n converges to Y with probability one,then Y n converges to Y in probability.Proof.Letǫ>0.We must showP(|Y n−Y|≥ǫ)=0limn→∞DefineE={ω:limY n(ω)=Y(ω)}n→∞Since Y n converges to Y with probability one,P(E)=1.LetE n=∩∞k=n{ω:|Y k(ω)−Y(ω)|<ǫ}As n gets larger,the intersection has fewer sets and so is larger.So E n is an increasing sequence.Ifω∈E,thenω∈E n for large enough n.SoE⊂∪∞n=1E nSince P(E)=1,this implies P(∪∞n=1E n)=1.By continuity of the probabil-ity measure,P(E n)1=P(∪∞n=1E n)=limn→∞Note that E n⊂{|Y n−Y|<ǫ}.So P(E n)≤P(|Y n−Y|<ǫ).Since P(E n) goes to1and probabilities are bounded by1,P(|Y n−Y|<ǫ)goes to1.So P(|Y n−Y|≥ǫ)goes to0.8.5Proof of central limit theoremWe give a partial proof of the central limit theorem.We will prove that the moment generating function of Z n converges to that of a standard normal.Let X n be an i.i.d.sequence.We assume thatµ=0.(Explain why we can do this.)Since the X n are identically distributed,they have the same mgf.Call it m(t).Som(t)=M X(t)=E[e tX n]n9As beforeZ n=S nσ√n,S n=nj=1X jNowM Zn (t)=E[exp(tS nσ√n)]=E[exp(tσ√n S n)]=M S n(tσ√n)Since the X j are independent,M Sn(t)= n j=1M X j(t)=m(t)n.So above becomesM Zn(t)= m(tσ√n) nNow we do a Taylor expansion of m(t)about t=0.We do the expansion to second order:m(t)=m(0)+m′(0)t+12m′′(0)t2+O(t3)We havem(0)=1m′(0)=E[X j]=0m′′(0)=E[X2j]=var(X j)=σ2 So the Taylor expansion ism(t)=1+12σ2t2+O(t3)Som(tσ√n) n= 1+12σ2t2σ2n+O(t3σ3n−3/2) n= 1+t22n+O(t3σ3n−3/2) nWe want to show that this converges to the mgf of the standard normal whichis exp(12t2).So we need to show the ln of the above converges to12t2.The ln10of the above isln(M Z n (t ))=n ln 1+t 22n +O (t 3σ3n −3/2) =n t 22n+O (t 3σ3n −3/2)+··· =t 22+O (t 3σ3n −1/2)+···which converges to t 2/2.The following theorem (which we do not prove)completes the proof of the central limit under an additional hypothesis about momemnt generating functions.Theorem 4.(Continuity theorem )Let X n be a sequence of random vari-ables,X a RV.Let F X n and F X be their cdf’s.Let M X n and M X be their moment generating functions.Suppose there is an a >0such that M X n (t )and M X (t )exist for |t |<a .Suppose that for |t |<a ,M X n (t )→M X (t ).Then for all x where F (x )is continuous,F X n (x )→F (x )The central limit theorem only requires that the random variables have a finite second moment,not that their mgf exist.To proof the CLT in this case we use “characteristic functions”instead of mgf’s.Definition 4.The characteristic function of a random variable X isφX (t )=E [e itX ]=E [cos(tX )+i sin(tX )]Since |cos(tX )|,|sin(tX )|≤1,the characteristic function is defined for all t for any random variable.For a continuous random variable with density f X (x ),φX (t )=∞−∞e itxf X (x )dx This is just the Fourier transform of f X (x )For a normal random variable the computation of the characteristic function is almost identical to that of11the mfg.Wefind for a standard normal random variable X thatφZ(t)= exp(−t2/2).If X and Y are independent,thenφX+Y(t)=φX(t)+φY(t)We also note that for a constant c,the characteristic function of cX is φcX(t)=φX(ct).Now let X have the Cauchy distribution.An exercise in contour integra-tion showsφX(t)=1π ∞−∞e ixt x2+1dx=e−|t|Now let X n be an iid sequence of Cauchy random variables.So if we let S n be their sum,thenφSn(t)=exp−n|t|This is the characteristic function of n times a Cauchy distribution.So1 nn j=1X jhas a Cauchy random variable.Note that the CLT would say1√nn j=1X jconverges to a normal.So instead of the CLT theorem we see that if we rescale differently then in the limit we get a Cauchy distributions.There are random variables other than Cauchy for which we also get convergence to the Cauchy distribution.Convergence of Binormial to Poisson????????????????8.6Proof of strong law of large numbersBorel Cantelli12。

Easily refutable subformulas of large random3CNF formulasUriel Feige Eran OfekDepartment of Computer Science and Applied MathematicsWeizmann Institute,Rehovot76100,Israelfeige,erano@wisdom.weizmann.ac.ilNovember9,2003AbstractA simple nonconstructive argument shows that most CNF formulas with clauses(where is alarge enough constant)are not satisfiable.It is an open question whether there is an efficient refutationalgorithm that for most formulas with clauses proves that they are not satisfiable.We present apolynomial time algorithm that for most CNF formulas with clauses(where is a large enoughconstant)finds a subformula with clauses and then proves that this subformula is not satisfiable(and hence that the original formula is not satisfiable).Previously,it was only known how to efficientlycertify the unsatisfiability of random CNF formulas with at least poly clauses.1IntroductionA CNF formula over variables is a set of clauses,each one contains exactly literals out of thepossible literals.A formula is satisfiable if there exists an assignment to its variables such that in each clause there is at least one literal whose value is true.The problem of deciding whether an input CNF formula is satisfiable is NP-hard,thus it is unlikely that there exists a polynomial time algorithm for it. Still,the NP-hardness of the3SAT problem does not rule out the possibility of designing a good heuristic for it.A heuristic for satisfiability may try tofind a satisfying assignment for an input formula,in case one exists.A refutation heuristic may try to prove that no satisfying assignment exists.In this paper we present an algorithm which tries to refute an input formula.The algorithm has one sided error,in the sense that it will never say“unsatisfiable”on a satisfiable formula,but for some unsatisfiable formulas it will fail to output“unsatisfiable”.It then follows that for a formula on which the algorithm outputs‘unsatisfiable’, its execution on is a witness for the unsatisfiability of.How does one measure the quality of a refutation heuristic?A possible test may be to check how good is the heuristic on a random input.But then,how do we generate a random unsatisfiable formula?To answer this question a short background on the satisfiability property of random CNF formulas is needed.The satisfiability property has the following interesting threshold behavior.Let be a random CNF formula with variables and clauses(each new clause is chosen independently and uniformly from the set of all possible clauses).As the constant is increased,it becomes less likely that is satisfiable,since the expected number of satisfying assignments reduces.In[FB99]it is shown that there exists such that for almost surely is satisfiable,and for,is almost surely unsatisfiable(for some which tends to zero as increases).It is also known that[KKL02],[JSV00].A possible method for generating random unsatisfiable CNF formulas is to take a random formula for. Such a formula is almost surely unsatisfiable.Similarly one can generate random satisfiable formula by setting.This threshold phenomena gives rise to two algorithmic tasks:(1)forgive an efficient algorithm that almost surelyfinds a satisfying assignment,(2)for give anefficient algorithm that almost surely proves that is unsatisfiable.Our work is related to the second task.Notice that for any,as is increased(for),the algorithmic problem of refutation becomes less difficult since we can always ignore afixed fraction of the clauses.The following question is still open:how small can be so that there is still an efficient algorithm which almost surely refutes random CNFformulas with clauses(may also be an increasing function of).A possible approach for refuting a formula is tofind a resolution proof for the unsatisfiability of.However,Chvatal and Szemeredi[CS88]proved that a resolution proof of a random CNF formula withlinear number of clauses is almost surely of exponential size.A result of a similarflavor for denser formulaswas given by Ben-Sasson and Wigderson[BSW01]who showed that a random formula with clauses (almost surely)requires a tree-like resolution proof of size.These lower bounds imply that finding a resolution proof for a random formula is computationally inefficient.A simple refutation algorithm can be used to refute random instances with clauses.This is done byfixing a variable,and taking all the clauses which contain it.Fixing to be true leaves about half of the selected clauses as a random CNF formula with clauses,which can be proved to be unsatisfied by a polynomial running time algorithm.The same can be done whenfixing to be false.A new approach introduced by Goerdt and Krivelevich in[GK01],gave a significant improvement andreduced the bound into clauses for efficient refutation of CNF formulas.This approach waslater extended in[FGK03]to handle also random CNF formulas with clauses.In[COGLS03],[FO03]it is shown how to efficiently refute a random CNF instances with at least clauses.Further motivation for studying efficient refutability of random CNF formulas is given in[Fei02].Thereit is shown that if there is no polynomial time refutation heuristic that works for most CNF formulas with clauses(where is an arbitrarily large constant)then certain combinatorial optimization problems(like minimum graph bisection,the-catalog segmentation problem,and others)have no polynomial time ap-proximation schemes.It is an open question whether it is NP-hard to approximate these problems arbitrarily well.Our refutation algorithm is based on similar techniques to those that appear in[FGK03],[COGLS03],[Fei02],but it has some advantages.Both the algorithm in[FGK03],[GL03]and our algorithm performeigenvalue computations on some random matrices derived from the random input formula.However,ourmatrices are much smaller(of order rather than),making the computational task easier.Moreover,ourstructure is simpler,making the analysis of our algorithm simpler,and easier to apply also to formulas withfewer clauses than those in[FGK03],[GL03].As a result of this simplicity,we can show that our algorithmrefutes formulas with clauses,whereas the algorithms given in[FGK03],[GL03]are claimed only torefute formulas with poly clauses respectively.In some other respects,our algorithm is more limited then the algorithms in[FGK03],[GL03].An algo-rithm is said to provide strong refutation if it shows not only that the input3CNF formula is not satisfiable,but also that every assignment to the variables fails to satisfy a constant fraction of the clauses.Our refuta-tion algorithm does not provide a strong refutation.It is plausible that the algorithms of[FGK03],[GL03](or a variant of them)does provide a strong refutation,though this issue is not explicitly addressed in[FGK03],[GL03].The ability to perform strong refutation is an important issue,and its relation to approx-imability is discussed in[Fei02].2Preliminaries2.1DefinitionsDefinition2.1.In the model there are vertices,each of the possible edges is chosen independentlywith probability.Definition2.2.In the model there are vertices,each of the possible directed edges is chosen independently with probability.Definition2.3.The is a random model for generating CNF formula.A formula is generated in the following way:each clause out of the possible clauses is chosen independently with probability of for some large enough constant.Although we concentrate on a specific random model for generating random formulas,our algorithm succeeds also on other related random models.For example,it is not hard to see that our algorithm works also if we generate a random formula by picking exactly clauses independently at random,with(or without)replacement.Details are omitted.Definition2.4().Let be two(multi)graphs on the same set of vertices.The graph is the multigraph with vertices whose edges are the union of(as multisets).2.2Efficient certification of a propertyAn important concept which will be frequently used is the concept of efficient certification.Let be some property of graphs/formulas or any other combinatorial object.An algorithm certifies the property if the following holds:1.On any input instance the algorithm returns either‘accept’or‘reject’.2.Soundness:The algorithm never outputs‘accept’on an instance which does not have the property.The algorithm may output‘reject’on an instance which has the property(the algorithm has one sided error).We will use certification algorithms on random instances of formulas/graphs taken from some probabil-ity space.We shall consider properties that are almost surely true for the random object taken from.A certification algorithm is considered good with respect to the probability space if it almost surely outputs ‘accept’on an input taken from.This property of the certification algorithm and the probability space is called Completeness.The computationally heavy part of our algorithm is certifying that two different graphs derived from the random formula do not have a large cut.One of these two graphs is random,and the other is a multigraph that is the union of6graphs,where each of these graphs by itself is essentially random,but there are correlations among the graphs.A cut in a graph is a partition of its vertices into two sets.The size of the cut is the number of edges with one endpoint in each part.A certification algorithm for verifying that an input graph with edges has no cut significantly larger than is implicit in[Zwi99].This algorithm is based on semi-definite programming;if the maximum cut in the input graph is of size at most, then the algorithm outputs a certificate that the maximum cut in is bounded by,where as.A computationally simpler algorithm can be applied if the graph is random.In [FO03]it is shown how to certify that in a random graph taken from the size of the maximum cut is bounded by,thus bounding the maximum cut by when is large enough.This is done by removing the vertices of highest degree from,and then computing the most negative eigenvalue of the adjacency matrix of the resulting graph.2.3An overview of our refutation algorithmOur algorithm builds on ideas taken from earlier work([FGK03],[Fei02],[FO03],[COGLS03],[GL03]). This section gives an informal overview of the algorithm at a fairly detailed level.Other sections of this manuscriptfill in the formal details.The input to the algorithm is a random3CNF formula with variables and clauses,where is a large enough constant.The algorithmfirst greedly extracts from a subformula.This is done as follows.We say that twoclauses are matched if they differ in theirfirst literal,but agree on their second literal and on their third literal.For example,the clauses and are matched.is constructed by greedily putting into pairs of clauses that form a match,until no further matches are found in.Let be the numberof clauses in.A simple probabilistic argument shows that we can expect. Moreover,is essentially a union of two random(but correlated)formulas and(each containing one clause from every pair of clauses that are matched in).Our algorithm will now ignore the rest of,and refute.As explained,is a union of two random formulas.Here we use an observation that is made in[Fei02],that we shall call the3XOR principle.The3XOR principle.In order to show that a random3CNF formula is not satisfiable,it suffices tostrongly refute it as a3XOR formula.Let us explain the terms used in the3XOR principle.A clause in a3XOR formula is satisfied if eitherone or three of its literals are satisfied.A strong refutation algorithm is one that shows that every assignmentto the variables leaves at least a constant fraction of the clauses not satisfied(as3XOR clauses,in our case).A proof of the3XOR principle is given in[Fei02].We sketch it here,and give it in more details inSection3.Observe that in a random formula every literal is expected to appear the same number of times,and if the number of clauses is large enough,then things behave pretty much like their expectation.As aconsequence,every assignment to the variables sets roughly half the occurrences of literals to true,androughly half to false.Hence every assignment satisfies on average3/2literals per clause.Moreover,thisproperty is easily certifiable,by summing up the number of occurences of the most popular literals.Given that every assignment satisfies on average3/2literals per clause,let us consider properties ofsatisfying assignments(if such assignments exist).The good option is that they satisfy one literal inroughly3/4of the clauses,three literals in roughly1/4of the clauses,and2literals in a negligiblefraction of the clauses.This keeps the average roughly at3/2,and indeed nearly satisfies the formulaalso as a3XOR formula,as postulated by the XOR principle.The bad option(which also keeps theaverage at3/2)is that the fraction of clauses that are satisfied three times drops significantly below1/4,implying that significantly more than3/4of the clauses are satisfied either once or twice,or in otherwords,as a3NAE(3-“not all equal”SAT)formula.But here,let us combine two facts.One is thatfor a random large enough formula,every assignment satisfies roughly3/4of the clauses as a3NAEformula.The other(to be explained below)is that there are known efficient algorithms for certifyingthat no assignment satisfies more than fraction of the clauses of a3NAE formula.Hence for arandom3CNF formula,one can efficiently certify that the bad option mentioned above does not occur.Having established the3XOR principle,the next step of our algorithm makes one round of Gaussianelimination.That is,under the assumption that we are looking for near satisfiability as3XOR(which issimply a linear equation modulo2),we can add clauses modulo2.Adding(modulo2)two matched clauses,the common literals drop out,and we get a clause with only two literals whose XOR is expected to be0,namely,a2EQ clause(EQ for equality).For example,from the clauses andone gets the clause.Doing this for all pairs of matched clauses in,we get a random2EQ formula.If was nearly satisfiable as3XOR,then must be nearly satisfiable as2EQ.But on the other,is essentially a random2EQ formula.For such formulas,every assignment satisfies roughly half the clauses.Moreover,there are known algorithms that certify this(to be explained shortly).Hence we can strongly refute as2EQ,implying strong refutation of as3XOR,implying strong refutation of as 3SAT,implying refutation(though not strong refutation)of as3SAT.Let us briefly explain here the major part that we skipped over in the description of our algorithm,namely,how to certify that a random2EQ formula is not satisfiable,and how to certify that a random3NAE formula is not satisfiable.In both cases,we reduce the certification problem to certifying that certain random graphs do not have large cuts,and then use the certification algorithms mentioned in Section2.2.(The principle of refuting random formulas by reduction to random graph problems was introduced in[FGK03].)To strongly refute random2EQ formulas,we negate thefirst literal in every clause,getting a2XOR formula.Now we construct a graph whose vertices are the literals,and whose edges are the clauses.A nearly satisfying2XOR assignment naturally partitions the vertices into two sides(those literals set to true by the assignment versus those that are set to false),giving a cut containing nearly all the edges.On the other hand,if the original2EQ formula was random,then so is the graph,and it does not have any large cut. As explained in Section2.2,we can efficiently certify that the graph does not have a large cut,thus strongly refuting the2XOR formula,and hence also strongly refuting the original2EQ formula.To strongly refute random3NAE formulas,we again condiser a max-cut problem on a graph(in fact, a multigraph,as there will be parallel edges)whose vertices are the literals.From each3NAE clause we derive three edges,one for every pair of literals.For example,from the NAE clause we get the edges,and.It is not hard to see that if a fraction of the3NAE clauses aresatisfied as3NAE,than a of the edges of the graph are cut by the partition induced by the corresponding assignment.Note that in our case(of that is the union of random and random )this graph is essentially a union of6random graphs:3graphs derived from the clauses of(one with edges derived from thefirst two literals in every clause,one with edges derived from the last two literals, and one from thefirst and third literal),and3graphs derived from.Hence it is not expected to have a cut containing significantly more than half the edges.One may certify that this is indeed the case either by using the algorithm of[Zwi99]on the whole graph,or by using the algorithm of[FO03]on each of the6 random graphs separately.Summarizing,our refutation algorithm extracts from a subformula(composed of matched pairs of clauses),checks that in almost all literals appear roughly the same number of times,derives from certain graphs on vertices and certifies that they do not have large cuts(e.g.,by computing the most negative eigenvalue of their adjacency matrices).The combination of all this evidence forms a proof that is not satisfiable.If is random and large enough(clauses),then almost surely the algorithm will indeed manage to collect all the desired evidence.3The refutation algorithmThe input formula is.For convenience we will assume that the clauses of appear in a random order and that the order of literals inside each clause is random(this assumption is reasonable because we can permute the clauses and the literals inside each clause before handling).We will use to denote a clause in which the second and the third literals are respectively and thefirst literal can be any literal. Let be a subformula of constructed in the following way:for every pair of literals we count the number of clauses in of the form.If this number is two or more,then we take into thefirst two appearances of these clauses(preserving the order of appearance).If the number of such clauses is one or less,we don’t add anything to.From each pair of clauses in we take thefirst one to the set and the second to the set.Each of is a random formula,though clauses in are not completely independent of each other:if a clause appears,then the clause cannot appear. From now own we will forget and concentrate in refuting.Before specifying the algorithm,we introduce additional notation and definitions which will ease the description of the algorithm.Definition3.1.Let be a3CNF formula with pairs of matched clauses.Then the following graphs,,,,and are defined as follows.:A graph whose vertices are the literals and whose edges are the following:each matched pair from ,say respectively,induces exactly one edge.Notice that we negate the literal which corresponds to the clause of.:A graph whose vertices are the literals and whose edges are the following:for each clause in we omit the-th coordinate and get a set of two literals which induces an edge of.:This graph is similar to but its edges are induced by.Definition3.2.A formula with clauses and variables is nearly-regular if the fraction of literals which are not regular is at most(a literal is regular if it appears in clauses),and the fraction of clauses containing not regular literals is bounded by.Definition3.3.A formula has the XOR property if the following holds:for every satisfying assignment at least fraction of the clauses are satisfied as XOR.Definition3.4.A graph is said to have a-cut if there is a partition of its vertices into two disjoint sets such that at least a-fraction of the edges of the graph connect the two sets.The number of matched pairs is denoted by.(In section2.3we used to denote the number of clauses in,and to denote the number of clauses in.From here on,may denote one of several different quantities,that will always be clear from the context.)In lemma A.2it is shown that almost surely .The refutation algorithm does the following steps(is some smallfixed constant):1.Certify that has the XOR property:if is satisfiable,then almost all its clauses aresatisfiable as XOR.This is done by certifying the XOR property for separately.If both have this property then also has this property.Specifically,verify the following(for ):(a)is nearly-regular(is some smallfixed constant).(b)Each of the graphs,,has a maximum cut bounded by.See theorem3.9fordetails.2.Certify that has a maximum cut with at most edges(the exact algorithm is shown inlemma3.14).See theorem3.9for details.The algorithm rejects if one of the above steps fails.We will show the following facts:(1)a CNF formula with clauses which satisfies the conditions(a),(b)(from step(1)of the refutation algorithm)has the XOR property;this is proved in lemma3.7.(2)The random graphs,,,,each has a maximum cut(almost surely)bounded by for small.This is done at theorem 3.8.(3)There is an efficient algorithm which certifies that each of,,,,has a maximum cut bounded by;this is shown in theorem3.9.(4)The above refutation algorithm almost surely accepts a random formula(corollary3.10).(5)A satisfiable formula will be rejected by the algorithm(corollary3.6).Theorem3.5.Let be a3CNF formula composed of pairs of matched clauses.Denote by be the graph induced by as described by the refutation algorithm.If the following conditions hold(for):1.has the XOR property.2.has no-cut.then is not satisfiable.Proof.We shall show that if is satisfiable and has the XOR property,then has a-cut. Consider the cut induced on the vertices of by a satisfying assignment(where in one side there are all the literals whose value is true and in the other side there are literals whose value is false).denotes the value of the literal induced by the assignment.By the XOR property of,all but fraction of the clauses of are satisfied as XOR clauses.Almost every pair of matched clauses induces an edge which crosses the cut:Let be a pair such that both and are satisfied as XOR. It holds that:modThus exactly one of the literals is true and the other is false(under the assignment),and the edge induced by this pair of clauses crosses the cut.It then follows that at least fraction of the edges are cut edges.Corollary3.6(Soundness).Let be a CNF formula.If the refutation algorithm accepts,then is not satisfiable.Lemma3.7(The XOR lemma).Let be a formula with clauses and variables,where. Let be the following projection graphs:has vertices associated with the literals of.The edges of are derived from by removing the-th coordinate from each clause of.Fix some.For (where is chosen such that)the following holds:if is nearly-regular and each has maximum cut bounded by then has the XOR property.Proof.The proof of this lemma appears at[Fei02];also a similar version of this lemma(for denser random CNF formulas)appears at[COGLS03].We repeat the proof for the sake of self-containment.We remind the reader that a literal is-regular if it appears in the formula times;in the following proof we will sometimes use the(shorter)term regular instead of regular while referring to literals.Since is-regular the fraction of not regular literals and the fraction of clauses containing not regular literals is bounded by.It holds that(in fact by increasing we can make the ratio between the two expressions arbitrarily small),thus the ratio between the number of regular literals and is as goes to infinity;the same holds for the ratio between the number of clauses containing only regular literals and.To ease the computations we will assume.Since we will perform only a constant number of arithmetic computations,the propagated error(induced by the ratios)can be made arbitrarily small by increasing.First we show that for every assignment all clauses are satisfied at most times on average. Every literal appears times thus the total number of satisfied literals is at most.Since the number of clauses is,every clause is satisfied at times on average.Let be a satisfying assignment.We will now show that the number of clauses satisfied by as AND is at least(a clause is satisfied as AND,if all of its literals are satisfied).Each has a maximum cut bounded by.Assume by contradiction that there are at most fraction of the clauses satisfied by as AND.It follows that there are at least clauses satisfied by as3(since satisfies all clauses).For each of the three graphs we focus on the cut induced by(on one side there are the satisfied literals,and in the other side the unsatisfied literals).We select a random edge from the union of the edges of the graphs by choosing a random clause and a random coordinate out of the three.If there are at least clauses satisfied as NAE then with probability of at leastthe induced edge is a cut edge in one of the graphs.It then follows that one of the graphs has a cut with at least edges,which leads to a contradiction.Thus the assumption that at most fraction of the clauses are satisfied as AND is false.It remains to show that all but fraction of the clauses are satisfied as XOR by.Denote by the fraction of clauses which are satisfied exactly once,exactly twice and exactly times re-spectively().We already know that;notice that,otherwise each clause will be satisfied more than on average.We have:substituting and with yields that and.In the following theorems.can be made arbitrarily small by increasing. Theorem3.8.Let be a random CNF formula chosen from(with large enough).Letbe subformulas derived from as described by the refutation algorithm.Then with high probability over the choice of,the subformula is nearly-regular,and none of the respective graphs,,and has a-cut.Proof.Let denote the number of matched pairs of clauses in.By lemma A.2almost surely .We will inspect the distribution of each of the random graphs.Out of the graphs it is enough to concentrate on since share the same distribution(recall that the clauses of are randomly permuted).To prove that is-regular,it is enough to show that in each of the graphs vertex is not regular,using the following reasoning(we condition on the event):(1)The expected fraction of not regular vertices(in each of the graphs)is bounded by.(2)The expected fraction of clauses containing not regular literals is bounded by.(3)Using Markov’s inequality we argue that with probability of at least the fraction of not regular literals is at most.The same argument is used for bounding the fraction of clauses con-taining not regular literals.The graph:Let be the number of pairs of clauses that are matched in.Each such pair induces a random edge in;the edges induced by different pairs are independent.We can think of as taken from the following distribution:The number of edges is chosen from the distribution(where).inde-pendent edges are uniformly selected among all possible(with repetitions).A standard large deviation bound(such as Chernoff’s)yields that the probability of afixed vertex to be not-regular is bounded by(conditioned on).A similar reasoning works also for the other graph models but we will omit the proof.The maximum cut is almost surely bounded by according to corollary3.12.The graph:In this multigraph for each couple there may be copies of the edge. For edges between are independent of edges between:the number of copies of the edge depends only on clauses of the form while the copies of depend on .By lemma A.1the probability for at least two appearances of clauses in is .Since copies of the(unordered)edge of are induced by pairs of clauses the form and pairs of clauses of the form,we conclude that i.e.is the sum of two random independent graphs.By lemma3.11almost surely each of the two graphs has a maximum cut bounded by,thus the maximum cut in is bounded by. Since(the total number of edges in)is almost surely(see lemma A.2)it follows that。

小学上册英语第三单元真题(含答案)考试时间：80分钟（总分:120）B卷一、综合题(共计100题共100分)1. 填空题：The ________ was a famous treaty that ended the Vietnam War.2. 听力题：The __________ is where most earthquakes occur.3. 听力题：The salad is very _______ (healthy).4. 选择题：What is the name of the president of the USA?A. KingB. Prime MinisterC. MayorD. President答案: D5. 填空题：The __________ (冷战) began after World War II.6. 选择题：What is the name of the famous ancient city in Greece?A. AthensB. SpartaC. DelphiD. Corinth答案:C. Delphi7. 选择题：What is 1 + 1?A. 1B. 2C. 3D. 4答案: B8. 填空题：The ________ has a long tail.9. 选择题：What do we call a strong windstorm that forms over the ocean?A. HurricaneB. TornadoC. CycloneD. Typhoon答案：A10. 填空题：The __________ (历史的影响) shapes our experiences.11. 听力题：They are ______ to the music. (dancing)12. 听力题：A __________ is a reaction that produces solid precipitates.13. 填空题：The _______ (猫) pounces on its toy.14. 填空题：I love playing ________ (角色扮演) games with my friends. We become different ________ (角色).15. 选择题：How many months are there in a year?A. TenB. ElevenC. TwelveD. Thirteen16. 填空题：My favorite book is _______ (傲慢与偏见)。