下载文档原格式
A.[4, 5, 6, lstA]B.[4, 5, 6, 1, 2, 3]C.[4, 5, 6, [1, 2, 3]]A.sumA()⽤循环⽅式求之和，sumB()⽤递归⽅式求B.默认情况下，倒数第⼆⾏被执⾏时如果输⼊较⼩的正整数如到之和C.默认情况下，倒数第⼆⾏被执⾏时如果输⼊较⼤的正整数如到之和D.默认情况下，⼀般说来，sumA()和效率⾼于sumB()第 5 题下⾯Python代码以递归⽅式实现字符串反序，横线处应填上代码是A.sReverse(sIn[1:]) + sIn[:1]B.sReverse(sIn[:1]) + sIn[1:]C.sIn[:1] + sReverse(sIn[1:])D.sIn[1:] + sReverse(sIn[:1])A.Hanoi(B, C, A, N-2)B.Hanoi(B, A, C, N-1)C.Hanoi(A, B, C, N-2)D.Hanoi(C, B, A, N-1)第 7 题根据下⾯Python代码的注释，横线处应填⼊( )。
A.isOdd isOdd(x)B.isOdd(x) isOddC.isOdd isOddD.isOdd(x) isOdd(x)A.checkNum()函数定义错误B.倒数第⾏代码将isOdd作为checkNum()参数将导致错误C.最后⼀⾏代码将Add作为checkNum()参数将导致错误D.倒数两⾏代码执⾏后都将没有错误第 10 题下⾯代码执⾏后的输出是（）。
A.4#3#2#2#4B.4#3#2#2#1#5C.4#3#2#1#2#4D.4#3#2#1#2#5第 11 题下⾯Python代码中的isPrimeA()和isPrimeB()都⽤于判断参数A.isPrimeA()的最坏时间复杂度是，isPrimeB()isPrimeB()B.isPrimeA()的最坏时间复杂度是，isPrimeB()isPrimeA()C.isPrimeA()的最坏时间复杂度是，isPrimeB()isPrimeB()D.isPrimeA()的最坏时间复杂度是，isPrimeB()isPrimeB()第 12 题下⾯Python代码⽤于归并排序，其中merge()函数被调⽤次数为（A. 0B. 1C. 6D. 7第 13 题在上题的归并排序算法中，代码Left, Right = mergeSort(listData[:Middle]), mergeSort(listData[Middle:])涉及到的算法有（）。
一、单选Unit 21. He C his daughter’s smile in this photograph.A. catchedB. tookC. capturedD. attracted2. We need to ensure a smooth D between the old system and the new one.A. transmissionB. translationC. transformationD. transition3. They complained about the A noise coming from the upstairs flat.A. excessiveB. extraC. massiveD. extensive4. The firm is B me for a decision on their offer.A. forcingB. pressingC. encouragingD. urging5. Children lose their B as they grow older.A. ignoranceB. innocenceC. confidenceD. experience6. That bit of form is for U.K. citizens—it doesn’t D to you.A. resortB. adaptC. referD. applyUnit 41. He has C about the proposals because they reduce workers’ rights.A. conceptionB. conservationC. reservationD. restoration2. The company A a management training program for its managers in different levels.A. initiatedB. foundedC. committedD. appealed3. A/An D number of members have called to complain.A. facialB. initialC. confidentialD. substantial4. Most countries have enthusiastically C the concept of high-speed railways.A. canceledB. eliminatedC. embracedD. conducted5. He is upset that his income was A compared with that of other chief executives.A. modestB. substantialC. considerableD. plentiful6. Doing voluntary work has added a whole new B to my life.A. illusionB. dimensionC. depressionD. conclusion二、选词填空Unit 11.He lives opposite to me.2.It’s traditional in America to eat turkey on Thanksgiving Day.3.He likes to give the impression that he’s terribly popular and has loads of friends.4.I felt physically and emotionally relaxed after the exam.5.Running can strengthen your heart and muscles, but conversely , it can also damageyour knee joints and the bones in your feet.6.We give preference to those who have worked with us for a long time.7.This will be a/an crucial decision for the education services because it sets thestandard for all future years.8.There have been several recent instances of planes taking off without adequatesafety checks.Unit 21.The invention of the wheel was a/an milestone in the history of man.2.Recorded highlights of today’s big football game will be shown after the news.3.Excessive drinking can lead to stomach disorders.4.We celebrated our 25th wedding anniversary in Florence.5.The article captured the mood of the nation at that time.6.This decision could have serious consequences for the future of the industry.7.He piloted the old lady through the crowd to her seat.8.An expert in any field may be defined as a person who possesses specialized skills and iscapable of rendering very competent services.Unit 31.The interview was sheer (十足的，完全的)torture from start to finish.2.The Japanese government has deliberately (故意的) sought to deceive the UnitedStates by false statements and expressions of hope for continued peace.3.Work permits were issued to only 5% of those who applied for them.4.I’m willing to dedicate myself wholeheartedly to ensure this country to becomestronger and more prosperous.5.She showed great perception in her assessment of family situation.6.It’s a small room, but the mirror creates a/an illusion of space.7.The students’ reactions were taken into account in shaping the final version of thedictionary.8.It is a universally acknowledged truth that a single man in possession of a good fortunemust be in want of a wife.Unit 41.They have come up with a plan to simplify the complex social security system.2.The government is committed to protecting the interests of tenants.3.The meeting is scheduled for Friday afternoon.4.Here are some good sources of information to assist you in making the best selection.5.Automation would bring a shorter, more flexible working week.6.Winning the match was just reward for the effort the team had made.7.To generate more money the sport needs to be more entertaining.8.He has expressed a/an desire to see you.三、介副词填空Unit 11.Sam couldn’t figure __out__ how to print a program until the teacher showed him how.2.I’m sick of having you hang __around___ with me.3.I think what appeals ____to____ me about his painting is the colours he uses.4.They met on a cruise, fell _____for______ each other, and married six months later.5.Many parents find it hard to relate ____to_______ their children when they are teenagers.6.She will end ____up____ penniless if she carries on spending as much as that.Unit 21.When I called on him, I found him poring __over_ Michael E. Porter’s On Competition.2.If you have applied ____to____ the Social Services to help with funding, you must be surethat you have their written approval before work starts.3.The decision should serve __as___ a warning to companies that pollute the environment.4.__In__ the eye of the law, theft is a less serious crime than handling stolen goods.5.The two countries were ____on_ the verge of war.6.Access __to__ up-to-date final information is important to the success in our investment. Unit 31.When our mind is indecisive, constantly wandering, and unable to concentrate, we willnever make it __to____ our destination.2.Worknet, which has a staff of 28 people, is dedicated __to____ assisting the community intheir search for employment or training.3.__As___one person puts it, we have enough problems getting young people to takemarriage seriously.4.He will appoint a five-member board that serves at his pleasure to overseedevelopment of the island for city use.5.The sound of the door closing deceived me _ _into____ thinking they had gone out.6.“Harmony” boasts strong Chinese characteristics, and expresses the traditional Chinesephilosophy __in____ pursuit of a balance between man and nature.Unit 41.What I am going to do in this lecture is to focus __on_____ something very specific.2.She had not committed anything __to_____ paper about it and soon she forgot it all.3.Take a clean sheet of paper and down the left-hand side make a list __of___ what you wantto buy.4.The meeting got bogged down with disputes about who was going to do what.5.I am not prepared to see children in this country having to settle __for___ a second classeducation.6.Miss Black was __at___ her best when she played the piano.7.The group meets __on___ a regular basis, usually weekly or biweekly.8.She is working __on___ the car.四、翻译（一）英译汉Unit 11.What mysterious force drives us into the arms of one person, while pushing us away from another who might appear equally desirable to any unbiased observer?到底是什么神秘力量使我们投入到某个人的怀抱，而不是在旁观者眼里同样可取的另一位？2.The mother has an additional influence on her sons: she not only gives them clues to what they will find attractive in a mate, but also affects how they feel about women in general.母亲对儿子有一个额外影响：她不仅暗示儿子在配偶身上什么是具有吸引力的地方，而且也会影响他们对女性的整体看法。
高三英语信息技术单选题50题6.She often _____ documents in the office software.A.editsB.makesC.createsD.designs答案：A。
本题考查动词在信息技术语境中的运用。
“edit”有“编辑”之意，在办公室软件中经常是编辑文档，符合语境。
“makes”通常指制作，范围比较宽泛，不如“edits”具体；“creates”强调创造新的东西，编辑文档不是创造新文档；“designs”主要是设计，与编辑文档的语境不符。
7.He _____ a new folder to store his files.A.buildsB.makesC.createsD.forms答案：C。
“create”有创建之意，创建新文件夹用“creates”比较合适。
“builds”通常用于建造较大的实体物体；“makes”制作的对象比较宽泛，不如“creates”准确；“forms”主要指形成某种形状或结构，不太适合创建文件夹的语境。
8.She _____ a file by mistake and had to restore it.A.deletedB.removedC.lostD.discarded答案：A。
“delete”表示删除，不小心删除了文件符合语境。
“removed”通常指移除某个物体，不一定是删除文件；“lost”是丢失，不一定是主动删除导致的；“discarded”侧重于丢弃不要的东西，不如“deleted”准确。
9.He _____ the file to another location.A.movedB.shiftedC.transferredD.carried答案：C。
“transfer”有转移、传送之意，把文件转移到另一个位置用“transferred”比较恰当。
“moved”和“shifted”比较笼统，没有“transfer”在信息技术语境中那么准确；“carried”通常指携带，不太适合文件转移的语境。
英语数字经济英语50题1. In the digital economy, we often talk about “big data”. What does “data” mean?A. 日期B. 数据C. 资料D. 图像答案：B。
“data”常见的意思是“数据”，A 选项“日期”是“date”，C 选项“资料”常用“information”，D 选项“图像”是“image”，所以这里应该选B。
2. The digital economy needs a lot of ______ to run smoothly.A. technologyB. technologiesC. technologicalD. technologically答案：A。
“technology”表示“技术”，是不可数名词，不能加s，C 选项“technological”是形容词“技术的”，D 选项“technologically”是副词“技术上地”，根据句意需要名词，所以选A。
3. In the world of digital economy, “e-commerce” is becoming more and more popular. What does “e-commerce” mean?A. 电子商务B. 电子商业C. 电子贸易D. 电子市场答案：A。
“e-commerce”常见的释义是“电子商务”，B 选项“电子商业”表述不准确，C 选项“电子贸易”通常用“e-trade”，D 选项“电子市场”一般是“e-market”，所以选A。
4. Digital economy brings many ______ to our life.A. convenientB. convenienceC. conveniencesD. inconvenient答案：C。
“convenience”是名词“便利”，是可数名词，many 后接可数名词复数，A 选项“convenient”是形容词“方便的”，D 选项“inconvenient”是“不方便的”，不符合句意，所以选C。
中考英语计算机编程单选题50题1.There are many programming languages. Java is a kind of _____.A.programming languageputer gameC.math subjectD.history book答案：A。
本题考查对编程相关术语的理解。
选项A“programming language”意为“编程语言”，Java 确实是一种编程语言；选项B“computer game”是“电脑游戏”；选项C“math subject”是“数学科目”；选项D“history book”是“历史书”。
2.In programming, a bug means _____.A.an insectB.a mistakeC.a flowerD.a book答案：B。
“bug”在编程中是“错误”的意思。
选项A“an insect”是“一只昆虫”；选项C“a flower”是“一朵花”；选项D“a book”是“一本书”。
3.A variable in programming can hold _____.A.numbers onlyB.letters onlyC.numbers and lettersD.just symbols答案：C。
变量在编程中可以存储数字和字母。
选项A 只说数字；选项B 只说字母；选项D 说只存储符号不准确。
4.When we write code, we use a(n) _____ to save and run it.A.textbookB.editorC.pencilD.eraser答案：B。
编写代码时，我们使用编辑器来保存和运行代码。
选项A“textbook”是“教科书”；选项C“pencil”是“铅笔”；选项D“eraser”是“橡皮”。
5.In programming, an algorithm is a set of _____.A.stepsB.picturesC.soundsD.colors答案：A。
B3U3知识清单Reading1.The first email was sent（send）in 1971 by Ray Tomlinson. It was a test message to himself. He chose the “@” sign to separate the username from the name of his machine.separate vt.使分开adj.独立的，分开的separate A from B 把A从B分离出来be separated from... 从...分离2.The first w ebcam was created by computer scientists at the University of Cambridge in 1991. What did it film? A coffee p ot outside the office. These scientists used the webcam to see from their desks whether the pot was empty to avoid wasted(waste) trips for coffee.webcam n. 网络摄像头[C]pot n. 壶，瓶，罐；锅[C]a coffee pot咖啡壶a pot of coffee 一壶咖啡pots and pans锅碗瓢盆a pot of jam一罐果酱3.Emojis, small digital images used(use)to express(express) ideas or feelings in electronic communication, were created(create) in 1999 in Japan.emoji n. 表情符号[C]4.The creator(create) was inspired(inspire) by Japanese c omics, street signs and Chinese characters.comic n. 连环漫画；喜剧演员[C] adj. 滑稽的；喜剧的→comedian n.喜剧演员→comedy n. 喜剧→反tragic adj.悲剧的→tragedy n.悲剧a comic actor喜剧演员comic strips四格漫画Chinese characters 汉字5.Soon emojis took off (突然大受欢迎，迅速流行) throughout the world. Oxford Dictionaries selected ..,the"Face with Tears of Joy" emoji, as the Word of the Year 2015.take off 突然大受欢迎，迅速流行The magazine took off after the interview. 那次采访之后这本杂志很快大受欢迎。
中考英语数字媒体单选题30题1.Which social media platform is known for short videos?A.WeChatB.InstagramC.LinkedInD.Twitter答案：B。
本题考查常见社交媒体平台的功能特点。
WeChat 主要是即时通讯，Instagram 以分享图片和短视频著名，LinkedIn 是职场社交平台，Twitter 主要是发布短消息。
2.People can share their daily lives on _____.A.FacebookB.QuoraC.PinterestD.Reddit答案：A。
Facebook 是人们可以分享日常生活的社交平台。
Quora 是问答平台，Pinterest 主要是图片分享，Reddit 是论坛式社交平台。
3.Which of the following is not a function of social media?A.Selling products directlyB.Making new friendsC.Sharing newsD.Playing games答案：A。
社交媒体的功能有结交新朋友、分享新闻、玩游戏等，一般不能直接销售产品。
4.You can follow celebrities on _____.A.SnapchatB.TumblrC.WeiboD.Yelp答案：C。
Weibo 上可以关注明星。
Snapchat 主要是分享瞬间照片和视频，Tumblr 是轻博客，Yelp 是点评平台。
5.What can you do on social media to express your opinions?A.Post commentsB.Send private messagesC.Add friendsD.Create groups答案：A。
在社交媒体上可以通过发表评论来表达自己的观点。
战略管理双语资料(共71页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--Chapter 1 Strateg ic Ma n a gem e nt a nd Str a tegic Com pe titiven e ss ................... 错误!未定义书签。
Management Process .............................................................................. 错误!未定义书签。
The Rational Model ....................................................................... 错误!未定义书签。
The critique of the rational model .................................................. 错误!未定义书签。
The New Competitive Landscape ........................................................... 错误!未定义书签。
Globalized Competition ................................................................. 错误!未定义书签。
Changes .......................................................................................... 错误!未定义书签。
I/O model of Above-average Returns ..................................................... 错误!未定义书签。
高三英语计算机语言单选题40题1. When you are programming, you often need to use a(n) ______ to store data.A. algorithmB. variableC. functionD. loop答案：B。
本题考查计算机语言中的常见词汇。
选项A“algorithm”意为“算法”；选项B“variable”指“变量”，在编程中用于存储数据，符合题意；选项C“function”是“函数”；选项D“loop”是“循环”。
2. In computer programming, a(n) ______ is a set of instructions that tells the computer what to do.A. codeB. scriptC. commandD. syntax答案：A。
选项A“code”指“代码”，是一组指令；选项B“script”通常指“脚本”；选项C“command”意为“命令”；选项D“syntax”表示“语法”。
本题强调的是一组指令，所以选A。
3. Which of the following is NOT a type of programming language?A. PythonB. ExcelC. JavaD. C++答案：B。
选项A“Python”、选项C“Java”和选项D“C++”都是常见的编程语言；选项B“Excel”是电子表格软件，不是编程语言。
4. The process of finding and fixing errors in a program is called ______.A. debuggingB. compilingC. optimizingD. documenting答案：A。
“debugging”意为“调试”，即查找和修复程序中的错误；“compiling”是“编译”；“optimizing”指“优化”；“documenting”表示“文档化”。
2021 年 12 ⽉真题（第⼀套）停⽌⽤⻝品标签贩卖恐慌 Stop the Food Label Fear-MongeringIn recent years, the has increased its use of labels.Whether the labels say 'non-GMO' or 'no sugar,' or 'zero carbohydrates', consumers are increasingly demanding more information about what's in their food.One report found that 39 percent of consumers would switch from the brands they currently buy to others that provide clearer, more accurate product information.Food manufacturers are responding to the report with new labels to meet that demand, and they're doing so with an eye towards giving their products an advantage over the competition, and bolstering profits.This strategy makes intuitive sense.If consumers say they want transparency, tell them exactly what is in your product.That is simply supplying a certain demand.food industry 近年来，⻝品⾏业增加了标签的使⽤。
Just how much does the Constitution（宪法）protect your digital data阅读理解答案阅读理解Just how much does the Constitution（宪法）protect your digital data The Supreme Court will now consider whether police can search the contents of a mobile phone without a warrant（授权令）if the phone is on or around a person during an arrest.California has asked the justices to restore the practice that the police may search through the contents of suspects’smartphones at the time of their arrest. It is hard, the state says, for judges to assess the implications of new and rapidly changing technologies.The justices would be careless if they followed California’s advice. They should start by rejecting California’s weak argument that exploring the contents of a smart phone is similar to say, going through a suspect’s wallet. The court has ruled that police don’t offend against the Fourth Amendment （修正案）when they go through the wallet, of an arrestee without a warrant. In fact, exploring one’s smartphone is more like entering his or her home. A smartphone may contain anarrestee’s reading history, financial history, medical history and comprehensive records of recent correspondence.Americans should take steps to protect their own digital privacy and should avoid putting important information in smartphones. But keeping sensitive information on these devices is increasingly a requirement of normal life. Citizens still have a right to expect private documents to remain private and protected by the Constitution’s prohibition on unreasonable searches.In many cases, it would not be very difficult for authorities to obtain a warrant to search through phone contents. They could still trump（打出王牌）the Fourth Amendment protections when facing severe and dangerous circumstances, such as the threat of immediate harm, and they could take reasonable measures to ensure that phone data are not deleted or altered while a warrant is on the way. The justices, though, may want to allow room for police to cite situations where they are entitled to more flexibility.Butthe justices should not swallow California’s argument whole. New technology sometimes demands fresh applications of the Constitution’s protections. Orin Kerr, a law professor, compares the explosion and accessibility of digital information in the 21st century with the establishment of automobile use as a digital necessity of life in the 20th. At that time, the justices had to explain new rules for the new personal domain（领域）of cars. Similarly, the justices must sort out how the Fourth Amendment of the Constitution applies to digital information now.73. The Supreme Court will work out whether, during an arrest, it is legal to ____________.A. search for suspects’mobile phones without a warrantB. check suspects’phone contents without being authorizedC. prevent suspects from deleting their phone contentsD. prohibit suspects from using their mobile phones74. The author’s attitude toward California’s argument is one of ____________.A. toleranceB. indifference C. disapproval D. carefulness75. The author believes that exploring one’s phonecontent is comparable to ____________.A. getting into one’s residenceB. handing one’s historical recordsC. scanning one’s correspondencesD. going through one’s wallet76. In Paragraph 4 and 5, the author shows his concern that ____________.A. principles are hard to be clearly expressedB. the court is giving police less room for actionC. phones are used to store sensitive informationD. citizens’privacy is not effectively protected77. Orin Kerr’s comparison is quoted to indicate that ____________.A. the Constitution should be carried out flexiblyB. New technology also requires reinterpretation of the ConstitutionC. California’s argument goes against principles of the ConstitutionD. Principles of the Constitution should never be altered答案解析：73-77 BCADB。
英语b和计算机考试试题及答案一、选择题（每题2分，共20分）1. What is the capital of France?A. LondonB. ParisC. RomeD. Berlin答案：B2. Which of the following is not a programming language?A. PythonB. JavaC. C++D. Photoshop答案：D3. The correct spelling of the word "independent" is:A. independantB. independantC. independantD. independent答案：D4. What does HTML stand for?A. Hyper Text Markup LanguageB. Home Tool Markup LanguageC. High Text Markup LanguageD. Hyper Text Markup Language答案：A5. Which of the following is a search engine?A. GoogleB. FacebookC. TwitterD. Instagram答案：A6. The function of a compiler in computer programming is to:A. Translate high-level code into machine codeB. Design the user interfaceC. Debug the codeD. Create the database答案：A7. In computer networking, what does IP stand for?A. Internet ProtocolB. Information ProtocolC. Internet ProviderD. Internet Processor答案：A8. The process of converting data into a form that can be easily transmitted over a network is known as:A. EncryptionB. CompressionC. SerializationD. Deserialization答案：C9. What is the primary function of an operating system?A. Running applicationsB. Managing hardware and software resourcesC. Storing dataD. Providing user interface答案：B10. Which of the following is a type of database management system?A. SQLB. HTMLC. CSSD. XML答案：A二、填空题（每题2分，共20分）1. The programming language that is often used for web development is ________.答案：JavaScript2. A ________ is a set of instructions that a computer can execute.答案：program3. The ________ is a protocol used for transmitting data overa network.答案：TCP/IP4. In computer science, ________ refers to the process of converting data from one format to another.答案：data conversion5. A ________ is a type of software that allows users to interact with the computer.答案：operating system6. The ________ is a type of memory that retains data even when the power is turned off.答案：non-volatile7. A ________ is a type of computer virus that replicates itself by sending copies to other computers.答案：worm8. The ________ is the primary storage for a computer's operating system, applications, and files.答案：hard drive9. A ________ is a set of rules that define how data is structured, constrained, and represented in a database.答案：schema10. In computer programming, ________ is the process of finding and fixing errors in the code.答案：debugging三、简答题（每题10分，共30分）1. Explain the difference between a compiler and an interpreter in programming.答案：A compiler is a program that translates high-level code into machine code that can be executed by a computer. It performs the translation once, and the resulting machine codecan be run multiple times without further compilation. An interpreter, on the other hand, translates and executes the code line by line during runtime, which means it does not produce a separate executable file.2. Describe the role of a firewall in a computer network.答案：A firewall is a network security system that monitors and controls incoming and outgoing network traffic based on predetermined security rules. It acts as a barrier between a trusted internal network and untrusted external networks, such as the Internet, and is used to protect a network from unauthorized access while allowing legitimate communications.3. What is the purpose of using version control in software development?答案：Version control is used in software development to track and manage changes to the source code over time. It allows multiple developers to work on the same project without overwriting each other's changes, provides a way to revert to previous versions if necessary, and helps in coordinating work among team members by maintaining a history of changes and who made them.。
英文版计算机试题及答案一、选择题（每题2分，共20分）1. Which of the following is not a function of an operating system?A. Process managementB. Memory managementC. Data storageD. File management2. In a computer network, what does the term "bandwidth" refer to?A. The width of the network cableB. The maximum rate of data transferC. The number of users connectedD. The speed of the network processor3. What is the primary purpose of a firewall?A. To prevent unauthorized access to a networkB. To encrypt dataC. To manage network trafficD. To store user passwords4. Which of the following is a type of software used for creating and editing documents?A. Spreadsheet softwareB. Database softwareC. Word processing softwareD. Graphics software5. What is the term used to describe the process of converting data from one format to another?A. Data migrationB. Data transformationC. Data conversionD. Data translation6. What does the acronym "CPU" stand for in computing?A. Central Processing UnitB. Central Processing UnitC. Computer Processing UnitD. Computing Processing Unit7. What is the function of a router in a network?A. To connect multiple networksB. To store dataC. To provide power to devicesD. To print documents8. What is the process of finding and fixing errors in software called?A. DebuggingB. PatchingC. UpdatingD. Patching9. Which of the following is a type of computer virus that replicates itself by attaching to other programs?A. TrojanB. WormC. RansomwareD. Spyware10. What is the term for the graphical representation of data on a computer screen?A. Data visualizationB. Data representationC. Data graphingD. Data mapping二、填空题（每题2分，共20分）1. The _________ is the primary memory used by a computer to store data and instructions that are currently being processed.2. A _________ is a type of software that allows users to create and edit images.3. The process of converting analog signals to digital signals is known as _________.4. A _________ is a collection of data stored in a structured format.5. The _________ is a hardware component that connects a computer to a network.6. In computer programming, a _________ is a sequence of statements that perform a specific task.7. The _________ is a type of malware that hides its presence and waits for a trigger to activate.8. A _________ is a type of software that is designed to protect a computer from unauthorized access.9. The _________ is the process of organizing and managing data in a database.10. A _________ is a type of software that allows users tocreate and edit spreadsheets.三、简答题（每题10分，共30分）1. Describe the role of a server in a computer network.2. Explain the difference between a compiler and an interpreter in programming.3. Discuss the importance of data backup and recovery in a computing environment.四、编程题（每题15分，共30分）1. Write a simple program in Python that calculates the factorial of a given number.2. Create a function in Java that takes an array of integers and returns the largest number in the array.答案：一、选择题1. C2. B3. A4. C5. C6. A7. A8. A9. B10. A二、填空题1. RAM (Random Access Memory)2. Graphics software3. Analog-to-digital conversion4. Database5. Network interface card (NIC)6. Function or procedure7. Trojan8. Antivirus software9. Database management10. Spreadsheet software三、简答题1. A server in a computer network is a powerful computer or system that manages network resources, including hardware and software, and provides services to other computers on the network, such as file storage, web hosting, and print services.2. A compiler is a program that translates source codewritten in a programming language into machine code that a computer can execute. An interpreter, on the other hand, reads and executes the source code line by line without the need for a separate compilation step.3. Data backup and recovery are crucial in a computing environment to prevent data loss due to hardware failure, software bugs, or malicious attacks. Regular backups ensure that data can be restored to a previous state in case of corruption or deletion.四、编程题1. Python Program for Factorial Calculation:```pythondef factorial(n):if n == 0:return 1 else:。
高三英语互联网技术单选题50题1. When you enter a website address in your browser, the part that comes after "www." is often the _______.A. IP addressB. domain nameC. protocolD. server name答案：B。
解析：本题考查互联网技术中的域名概念。
A选项IP 地址是互联网协议地址，是设备在网络中的标识，与网址中“www.”后的部分概念不同。
B选项域名是网站的标识，通常位于“www.”之后，符合题意。
C选项协议是网络通信的规则，与网址的这部分内容无关。
D选项服务器名称主要指服务器的标识，并非网址中“www.”后的部分内容。
2. The _______ is a set of rules that govern how data is transmitted over the Internet.A. domainB. URLC. protocolD. cache答案：C。
解析：本题考查网络协议的概念。
A选项域名是网站的标识，不是管理数据传输的规则。
B选项统一资源定位符（URL）是用来定位网页的地址，不是数据传输规则。
C选项协议是管理网络上数据传输的一套规则，符合题意。
D选项缓存是存储临时数据的地方，与数据传输规则无关。
3. Which of the following is an example of a top - level domain?A. comB. wwwC. httpD. server答案：A。
解析：本题考查顶级域名的概念。
A选项“com”是一种常见的顶级域名，用于商业网站等。
B选项“www”是万维网的标识，不是顶级域名。
C选项“http”是超文本传输协议，不是顶级域名。
D选项“server”是服务器的意思，不是顶级域名相关概念。
中考英语大数据的分析方法与应用单选题40题1.Big data is composed of a large number of ______.rmationsB.dataC.datumD.message答案：B。
information 是不可数名词，没有复数形式；data 是不可数名词，通常用复数形式；datum 是单数形式，意思是数据，资料；message 是信息的意思，与大数据的组成不符。
大数据是由大量的数据组成。
2.In big data analysis, we often deal with ______ of different types.A.elementB.elementsC.factorD.factors答案：B。
element 是元素的意思；elements 是复数形式；factor 是因素的意思；大数据分析中，我们经常处理不同类型的元素，element 是可数名词，这里要用复数形式。
3.The core of big data is the processing and analysis of massive ______.A.datumB.datarmationD.message答案：B。
datum 是单数形式；data 是不可数名词，大数据的核心是对大量数据的处理和分析；information 和message 与题干中的核心不符。
4.Big data contains a wealth of ______.A.detailB.detailsrmationD.datums答案：C。
detail 是细节的意思；details 是复数形式；大数据包含丰富的信息，information 是不可数名词；datum 的复数形式是data。
5.In the field of big data, ______ is crucial for decision-making.A.datarmationC.elementD.factor答案：A。
python初中竞赛试题及答案一、选择题（每题3分，共30分）1. 以下哪个选项是Python中用于定义函数的关键字？A. defB. functionC. declareD. define答案：A2. 在Python中，以下哪个选项是正确的字符串定义方式？A. 'Hello, World!'B. "Hello, World!"C. Both A and BD. None of the above答案：C3. Python中，以下哪个选项是正确的注释方式？A. // This is a commentB. # This is a commentC. /* This is a comment */D. Both A and C答案：B4. 在Python中，以下哪个选项是正确的列表定义方式？A. [1, 2, 3]B. (1, 2, 3)C. {1, 2, 3}D. A and B答案：D5. 下列哪个选项是Python中的True值？A. 0B. 1C. NoneD. False答案：B6. 在Python中，以下哪个选项是正确的字典定义方式？A. {key: 'value'}B. [key: 'value']C. (key: 'value')D. {key: 'value', 'key2': 'value2'}答案：D7. Python中，以下哪个选项是正确的条件语句？A. if condition:B. if conditionC. if condition thenD. All of the above答案：A8. 在Python中，以下哪个选项是正确的循环语句？A. for item in range(5):B. for item in 5:C. while condition:D. Both A and C答案：D9. 下列哪个选项是Python中用于创建空列表的语法？A. list = []B. list = list()C. Both A and BD. None of the above答案：C10. Python中，以下哪个选项是正确的元组定义方式？A. (1, 2, 3)B. [1, 2, 3]C. {1, 2, 3}D. A and B答案：A二、填空题（每题4分，共20分）1. 在Python中，使用________关键字可以定义一个类。
Digital Watermarking of Chemical Structure SetsJoachim J.Eggers,W.-D.Ihlenfeldt,and Bernd GirodTelecommunications Laboratory,University of Erlangen-NurembergCauerstr.7/NT,91058Erlangen,Germany,eggers@LNT.deComputer Chemistry Center,University of Erlangen-NurembergN¨a gelsbachstr.25,91052Erlangen,Germany,wdi@ccc.chemie.uni-erlangen.de Information Systems Laboratory,Stanford UniversityStanford,CA94305-9510,USA,girod@ 4th Information Hiding Workshop Pittsburgh,PA,USA 25-27April,2001Abstract.The information about3D atomic coordinates of chemical structuresis valuable knowledge in many respect.For large sets of different structures,thecomputation or measurement of these coordinates is an expensive process.There-fore,the originator of such a data set is interested in enforcing his intellectualproperty right.In this paper,a method for copyright protection of chemical struc-ture sets based on digital watermarking is proposed.A complete watermarkingsystem including synchronization of the watermark detector and verification ofthe decoded watermark message is presented.The basic embedding scheme,de-noted SCS(Scalar Costa Scheme)watermarking,is based on considering water-marking as a communications problem with side information at the encoder.1IntroductionChemical structures are inherently three-dimensional,although most structure databases store them only asflat graphs.For many scientific studies,for example the development of drugs,the3-D structure is a major factor determining the application potential of a compound.It is possible to determine3-D atomic coordinates by experimental tech-niques,but this is very expensive.As an alternative,computational methods of various precision levels exist which take a structure graph or very rough3-D structure approx-imation as input and compute3-D atomic coordinates.For large datasets containing hundreds of thousands of molecules,quantum-chemical or fully optimizing force-field methods are not usable because they are too computationally expensive.Expensive opti-mizations can largely be avoided by model builders which employ complex rule-driven heuristics.The development of such programs is difficult,and represents a significant investment.Consequently,these programs are expensive when bought commercially, and coordinate sets,which are needed to isolate functional principles common among compounds with similar biological activity,represent a tangible value,even if the un-derlying structures are in the public domain.Due to the value of computed structure data,the originator is interested in enforcing the copyright of the data.Thus,robust labeling and identification of structure data is desired.Here,digital watermarking of the molecule structure data is investigated as one method for such labeling and identifi-cation.The intellectual property of the data set resides only in the atomic coordinates. Taking into account the limited precision of the model builder,a variation of the co-ordinates is acceptable and can be used for watermarking purposes.Given the smallsize of typical records for one structure,it is certainly not possible to robustly mark every record,but this is not necessary.We are mainly interested in identifying the ori-gin of large data sets,e.g.,including100,000-200,000structures.Resistance against tampering by adding small amounts of random jitter to the coordinates,in addition to resistance against rotations and translations,is desirable.A more comprehensive list of possible attacks is given in Section4.1.Digital Watermarking has been investigated intensively during the last years in the context of multimedia data,e.g.,audio,image or video data.Most blind watermarking techniques,where the watermark detector has no access to the original data,are based on spread-spectrum techniques,but recently much more powerful techniques have been proposed.One such method is called SCS(Scalar Costa Scheme)watermarking.SCS watermarking is appropriate for many different data characteristics,and thus is used here for embedding watermarks into the molecule data.In Section2,the basic principles and design criteria for SCS watermarking are re-viewed.Next,the problem of detecting the existence of a SCS watermark is discussed in Section3.In Section4,the specific system design for SCS watermark embedding into and detection from the chemical structure data is described.The performance of the proposed scheme is investigated experimentally,and simulation results are presented in Section5.2SCS WatermarkingWe consider digital watermarking as a communication problem.The watermark en-coder derives from the watermark message(sometimes also called“payload”)and the host data an appropriate watermark sequence which is added to the host data to produce the watermarked data.must be chosen such that the distortion betweenand is negligible.Next,an attacker might modify the watermarked data into data to impair watermark communication.The attack is only constrained with respect to the distortion between and.Finally,the decoder determines from the received dataan estimate of the embedded watermark message.The encoder and decoder must be designed such that with high probability.In blind watermarking schemes, the host data are not available to the decoder.The codebook used by the watermark encoder and decoder is randomized dependent on a key to achieve secrecy of wa-termark ually,a key sequence is derived from to enable secure watermark embedding for each host data element.Here,,,,and are vectors,and ,,,and refer to their respective th elements.Fig.1depicts a block diagram of blind watermark communication,where an attack by additive white Gaussian noise(AWGN)is assumed.The depicted scenario can be considered communication with side information about the host signal at the encoder. For this scenario,Costa[3]showed theoretically that for a Gaussian host signal of power,a watermark signal of power,and AWGN of power the maximum rate of reliable communication(capacity)is,independent of .The result is surprising since it shows that the host signal need not be considered as interference at the decoder although the decoder does not know.IIcheme)[4].Note that SCS is very similar to Costa’s original scheme,except for the suboptimal scalar quanitzer.Thewatermark message is encoded into a sequence of watermark letters,where in case of binary SCS.Note that this encoding process is usually divided into three steps.First,is represented by a vector with binary elements.Second,is encoded into by a binary error correcting code.Finally,is mapped on by selection or repetition of single coded bits so that each of the watermark letters can be embedded into the corresponding host element.The embedding rule for the th element is given by(1) where denotes scalar uniform quantization with step size,and is the er-ror of subtractive dithered quantization.The key is a pseudo-random sequence with .The upper plot of Fig.2depicts one period of the PDF of the sent elements conditioned on the sent watermark letter and.The described embedding scheme depends on two parameters:the quantizer step size and the scale factor. Both parameters can be jointly optimized to achieve a good trade-off between embed-ding distortion and detection reliability for a given noise variance of an AWGN attack. Optimal values for and are given in[4].In general,if accurate statistical models of the host data are unavailable,and a MSE distortion measure is used,and can be designed for an AWGN attack with a specific watermark-to-noise power ratio(WNR). Note that this heuristic is only useful if a potential attacker does not have an accurate model for the host signal either.At the decoder,the received data is demodulated to obtain the data.The demod-ulation rule for the th element is(2)where.should be close to zero if was sent,and close to for .The lower plot in Fig.2shows the PDF of the demodulated elements afterIIIp (s |d )sp (y |d )yFig.2.One period of the PDFs of the sent and the received signal for binary SCS (=1,WNR dB,,).The filled areas represent the probability of detection errors assuming was sent.The dotted line in the lower plot depicts the PDF when detecting with a wrong key .AWGN attack conditioned on the sent watermark letter.can be computed numerically as described in [4].In case of using an incorrect key at the receiver,the distribution of will be uniform for any possible .This is indicated by the dotted line in the lower plot of Fig.2.The performance of SCS watermarking is discussed in detail in [4,5].It can be shown that for a large range of different WNRs SCS watermarking is superior to com-mon blind spread-spectrum watermarking schemes since spread-spectrum watermark-ing suffers from large host signal interference.Note that the resiliency of SCS against AWGN attacks is independent from the host distribution.This property is particularly important for the application at hand,since the molecule coordinates of chemical struc-tures do not have a smooth distribution,e.g.,Gaussian or Laplacian,which is usually assumed in the design of detectors for spread-spectrum watermarks.It was also shown that at low watermarking rates,Spread Transform (ST)SCS watermarking is superior to SCS watermarking with simple repetition coding [5].ST watermarking was originally proposed by Chen and Wornell [2]to improve binary dither modulation watermarking.In ST watermarking,the watermark is not directly embedded into the host signal ,but into the projection of onto a random sequence of length .Any noise orthogo-nal to the spreading vector does not impair watermark detection.Thus,an attacker,not knowing the exact spreading direction ,has to introduce much larger distortions to im-pair an ST-SCS watermark than a simple SCS watermark.For AWGN attack and MSE distortion measurements,doubling the spreading length gives an additional power ad-vantage of 3dB for the ST-SCS watermark.Of course,this gain in detection reliability comes with a decrease of the watermark rate.In general,a spread transform of lengthIVrequires-times more data elements for watermark embedding.The optimal spreadinglength depends on the strength of attacks to be survived.3Verification of Decoded Watermark InformationSo far,watermarking was considered as a communication problem where at the water-mark decoder a watermark message is received assuming that a watermark is embed-ded with the key.However,in many watermarking applications the detector has todecide whether a watermark with key is embedded in the received data at all.Note that this problem differs somewhat from the communication problem.For SCS watermarking we do not distinguish between the following cases:–receiving non-watermarked data,–receiving data that is watermarked with a different watermarking technique,–receiving data being SCS-watermarked with a different key than the key.This is justified by the host signal independent nature of SCS watermark detection and the use of a key sequence with being uniformly distributed in.Subsequently,we only distinguish between watermark detection from data watermarked with keyand from data not watermarked with key.We assume that SCS watermarking was designed to communicate the message asreliably as possible via the watermarking channel.However,trying to detect an SCS wa-termark using a wrong key,leads to demodulated data that is uniformly distributed within as indicated by the dotted line in Fig.2.Thus,the decoded water-mark message will be a random bit sequence with. The problem of deciding whether is a valid watermark message or just a random bitsequence can be formulated as a hypothesis test between–hypothesis:no watermark message is embedded in with key,and–hypothesis:the watermark message is embedded in.In general,both hypotheses cannot be separated perfectly.Thus,we have to trade offthe probability of accepting when is true(false positive)and the probability of accepting when is true(false negative).Here,we devote a sub-vector of length of the watermark message for ver-ifying the validity of a received watermark message.We compare two methods to decide between and using the verification bit vector.In ourfirst approach, called method A,is equal to thefirst bits of and error correction coding of is such that thefirst bits of the coded watermark message are independent from the remaining watermark message bits.When detecting an SCS watermark letter from a data element where the embedded letter is one of the coded verification bits,the probabilities for receiving a demodulated value depending on hypothesis or are given asLet denote the index set of all data elements with embedded coded verification bits. Due to the independent identically distributed key sequence,the respective probabil-ities for detection from all data elements with index are given by(5)(6) Applying Bayes’solution to the hypothesis test with equal a priori probabilities and equal costs for both hypotheses,is accepted if4.1Attacks on Chemical Structure DataThe type of attacks which can be envisioned for structure data sets is notably different from those applicable to audio-visual data and similar,classical areas of digital wa-termarking.First of all,raw watermarked structures can again be subjected to various different energy minimization procedures,including the algorithm which was initially used to generate the data,essentially re-computing the protected information.Protection against this type of attack is not possible.The initial information about the structural identity needs to be contained in the datafile and can be used as basis for any further computation.However,we assume that no unlicensed copies of the software used to generate the original protected data are in circulation.Further,the computation time for larger datasets is often significant.Depending on the type of algorithm used and the size of the dataset,it can be up to several CPU months.Thus,simple re-generation of the data is often not a feasible approach.Attacks to remove or dilute the watermark are then limited to a small set of general,computationally inexpensive operations.These include:–Removal of data from the original dataset,or injection of additional structures that are not watermarked,but possess coordinates from other,unmarked or differently marked sources.–Re-ordering of the individual records in the dataset.–Re-ordering of atoms and bonds in the structure records.–Global3-D transforms.Rotating or shifting the structures in3-D space does not change their usability,since the intermolecular distances,angles and torsions define the characteristics of a molecule,not its orientation in3-D space.–Variation of structure notation.In some cases,structural features can be represented by different notational conventions without changing the identity of the structure.For instance,in a common format aromatic systems are represented as Kekul´e sys-tems.The sequence of single and double bonds can be re-arranged without chang-ing the structure.These are comparatively simple operations,and all identification algorithms which use the structure as access key or generate canonic orderings of atoms will have to cope with this variability.–Removal of atoms from structures.This operation is clearly a major modification of the structure,and the only case where the data retains at least a part of its usefulness is the global removal of hydrogen atoms.4.2Initial Considerations for the System DesignIn general,a single structure does not contain enough data for an entire watermark message.Thus,the watermark message is distributed over several molecule structures, and watermark detection is only possible when several,perhaps modified,structures are available.The illegal use of single molecules cannot be proven,however,heavy illegal use of a large amount of the structure data should be detectable.The watermark detector must have some information about the exact location of embedded watermark bits even after data re-ordering attacks as mentioned above.This problem is related to the well-known synchronization problem of watermark detectors.VIIOur system design is such that perfect synchronization of the watermark detector is always ensured.The required algorithms are described below.The watermark message is represented by a vector of binary elements().can include different information,e.g.,an identifier of the copyright holder, a verification bit vector,and/or the date of computation of the molecule data.Fur-ther,the watermark embedding is dependent on a key,which is only known to the copyright holder and perhaps a trusted third party.The detection reliability may be improved by error correction codes.Thus,is en-coded into a binary vector of length.The influence of different error correction codes is investigated experimentally in Section5.Note that only some of the en-coded watermark bits in will be embedded into one molecule.Thus,decoding of must be possible even if some of the encoded bits are not available from the data given at the watermark detector.To solve this problem,as much watermark information as possible is collected from each molecule,and this information has to be combined correctly to decode the watermark message.4.3Structure Normalization and Hash ComputationIn an attempt to embed or detect a watermark,the structure needs to be normalized and identified.Only parts of the encoded watermark message are embedded into each single structure(see Section4.4).The specific message part to be embedded is determined by a hash code generated from the structure at hand.The hash code depends solely on the structure description.Thus,the watermark encoding and decoding process is independent of the order of structures in a large dataset.Also,insertion of unmarked records and deletion of marked records can be accepted to a comparatively high extent, since the additional or missing structures will only reduce the detection reliability,but no synchronization problems will ensue.Hash codes for chemical structures being invariant to the operations mentioned in Section4.1,exhibiting good randomness and negligible correlation in all bits,and do not generate hash code collisions for closely related structures,are not trivial,and have been studied extensively in chemoinformatics.We are using a state-of-the-art64-bit hash code[7]which has some proven advantages over earlier attempts.Since the hash code depends on the hydrogen addition status,we always add a standard hydrogen set to the structure before computing the hash.If the hydrogen atoms are present,this is a null operation,otherwise new atoms are added with undefined or at least unmarked coordinates.These atoms,if added,will reduce detection reliability,but will ensure that the original structure hash code is regenerated and the original canonical atom order can be obtained.Once the encoded message part has been identified with help of the hash code,the next preparation step is to move the structure to an unique3-D orientation and generate a canonical ordering of the atoms.The canonical order of the atoms is determined by a symmetry-breaking sphere-expansion process.We use an adapted version of the Unique SMILES algorithm by Weininger et.al.[8].This method is fast and exact for practically relevant structures.A few errors in re-establishing the precise atom order in highly symmetrical structures can be tolerated.We have enhanced the original algorithm toVIIIinclude hydrogen atoms(whose coordinates are important)and to break some additional symmetric cases in a deterministic fashion.4.4Watermark Embedding into a Single Molecule StructureThe embedding of encoded watermark bits from in the th structure is consid-ered.A block diagram of the embedding scheme is depicted in Fig.3.First,a canonic representation of the structure is obtained as described above.Next,the host data vec-tor is extracted(see also Section4.7).Here,it is assumed that elements are extracted from the structure,and the elements of are scaled such that the water-mark can be embedded with a variance.,a pseudo-random key vector with elements is required to hide the embedded watermark to malicious attackers.The pseudo-random vectors,,and must be perfectly reconstructible at the watermark detector and should not be known to unauthorized parties.Thus,the64bit hash value of the structure is taken as seed for a cryptographic secure random number generator which is used to compute,andfrom this hash value dependent on the key of the copyright holder.In the current implementation a pseudo-random number generator based on DES encryption is used.The watermark letters are embedded into and the watermarked vector is obtained.Finally,the inverse spread transform is applied to obtain which is combined with the unmodified structure information to synthesize the watermarked molecule structure(10)These probabilities are collected in the vector.The required conditional proba-bilities and depend on the used watermarking scheme,but also on possible attacks.We designed our scheme for an AWGN attack of a certain noise variance,e.g.WNR dB.This heuristic is useful since up to now little about possible statistical attacks on the watermarked structure data is known.The vectors and are the result of the detection process for the molecule.4.6Joint Watermark Detection from Several MoleculesAssume that structures are received,with.The vectors and of length are derived as described above from each received structure.Further, we assume that the attack on the embedded watermark is memoryless,that is all demod-ulated watermark letters are statistically independent.Thus,the probabilitythat the th coded watermark bit is1,is given by(11)XThe quality of a3-D structure dataset is measured by the energy(enthalpy of for-mation)of the conformers.Good coordinate generators will display a good balance be-tween execution speed and conformer energy.The quality of a dataset can be checkedby comparing the energy of the dataset structures to the energies obtained by using amore computationally expensive method to optimize the3-D structures.Our primary test dataset was generated by the3-D coordinate generator CORINA[6]which is veryfast and employs only a low level of theory(rule-based initial coordinate generationand pseudo-forcefield energies for optimization).Since the testing of the acceptability of the watermarked structures requires a better level of theory than the original gen-erator,we used the AM1implementation of the V AMP package[1]which has been successfully used to process the same data set in a very expensive computational effort.The acceptable level of distortion of the original coordinates depends on the preci-sion of the original results.For CORINA coordinates,a change of2-3%of the structureenergy is tolerable.For an AM1data set,less than1%would be acceptable.For the CORINA dataset,we measured the compound energy before and after watermarkingby performing a single-point AM1computation which will not change and re-optimize the coordinates but only compute the energy of that coordinate set.In the current im-plementation,the modification of the atomic coordinates does not take into accountthe atomic environment at all.However,not all distortions of the structures lead to the same energy change.Thus,improved allocation of the watermark power to differentcoordinates should be investigated in the future.5Performance EvaluationThe described system for watermarking of chemical structure sets involves many differ-ent parameters,like the error correction code,the spread transform length,the water-mark message length,the parameter and for SCS watermarking,and the choiceof verification bits.A detailed discussion of all parameters is beyond the scope of this paper.Here,we consider a watermark message offixed length bits(equiv-alent to12ASCII characters).The parameter and were designed for an AWGNattack with WNR dB.Thus,the SCS scheme was optimized for an AWGN at-tack where the power of additive noise is twice as large as the watermark power .Most of the experiments discussed below were performed on synthetic data since many simulations are required to measure low error probabilities.Nevertheless,some simulations results for chemical structure sets will be discussed,too.5.1Required Amount of Received Data ElementsThe watermark bit error probability was investigated experimentally for different amounts of received data elements.In practice,reliable detection from as few data elements as possible is desired.We restrict the discussion to an AWGN attack with WNR dB.Rate1/3convolutional codes(CC)with memory andwere used to encode all watermark bits into the coded bit vector with length and,respectively.XIIrandom data elements were chosen as host signal.This data was transformed into the spread transform domain where the projected data has el-ements.Note that for.For each element in,one bit of the en-coded watermark message was randomly selected and embedded.Simulations with 20000random watermark messages were performed so that bit error probabilities about can be measured reliably.Fig.5shows the measured bit error probabilities for CC with and,and spread transform lengths and. Obviously,the scheme with and performed best.Only2000data elements are required to achieve.This corresponds to a watermark rate of aboutbit/element.About1000more data elements need to be received when using the less complex convolutional code with.Another500more data elements are required when leaving out the spread transform().Fig.5.Measured bit error probabilities for receiving96watermark message bits af-ter AWGN attack with WNR=-3.0dB. The watermark message was encoded with a rate1/3convolutional code with differ-ent memory length.Simulation results for spread transform lengths andare shown.Fig.6.False positive and false negative error probabilities for watermark verifica-tion.Two methods using15verification bits are compared.The watermarked data is at-tacked by AWGN with WNR dB.Note that the considered detection case is different from detection after a simple AWGN attack.The detection performance is impaired also by the randomness with which certain data elements are received.Simulation results show that lower error prob-abilities could be achieved when the number of embedding positions would be identical for all coded bits.However,in the application at hand,it is impossible to ensure that the watermark detector receives all watermarked data elements.5.2Verification of Decoded WatermarkTwo methods for verifying the validity of a received watermark message were proposed in Section3.Here,simulation results for both methods are compared.Fig.6shows the measured false positive and false negative probability for a verification bit vector of length.The watermark message was embedded with a rate CC with memory length.XIIIThe detection of200000random watermark messages was simulated and different amounts of received data was considered.The SCS parameter and channel noise werechosen as in the previous subsection.Hypothesis was valid in half of the cases,thus the error probabilities were estimated from100000decisions.For method B,a false positive probability can be expected.This value is verified bythe simulation results shown in Fig.6.The false negative error probability of method B depends on the bit error probability which decreases for an increased number of received data elements.Fig.6shows that of method B also decreases slowly withthe number of received data elements.Contrary,for method A the error probabilities and are almost identical when receiving few data elements.For an increasednumber of received data elements more false negative errors than false positive errors occur.Method B is superior with respect to the false positive rate when detecting from few data elements.However,the overall error probability is lower for method A.Notethat for method A it also possible to achieve lower false positive rates by increasing the decision threshold which was0.5in(7).Of course higher false negative rates have to be accepted in such a case.5.3Perfect Attack on Parts of the DataIt is likely that an attacker has perfect knowledge about the original data for some part of the data set.In this case,the attacker simply replaces the watermarked data by theoriginal data,thus erasing the watermark from the specific data elements.In general we found that reliable watermark detection can be achieved even for a substitution of80%of the watermarked data elements.However,this is only possible when many data elements are available at the decoder.Thus,it is worth to select for the watermarking process only data elements which are unlikely to be known by an attacker.The disturb-ing influence of data replacement can be prevented this way.5.4Simulations with Example Molecule DataPreliminary experiments with example molecule data were conducted.The host vectorswere composed by all atom coordinates of one molecule structure.The coordinate values were scaled by a factor of1000such that a watermark of power can be embedded.For this setting the AM1energies in a200-structure test set were changedby less than0.3%on average,without producing outliers with unacceptable energies (more than1.5%energy increase,corresponding to unusable structures).25%of the structures were actually lower in AM1energy after watermarking,demonstrating the imperfectness of the CORINA optimizer.The watermark was detectable on this comparatively small dataset with near100%confidence even after performing the following set of operations:Delete10random structures,add10similar structures without a watermark,re-compute unmarked coor-dinates for10random molecules,shuffle the sequence by moving50random structures into different slots andfinally randomly rotate and translate all molecules.The algo-rithm proved to be very robust against this set of operations which we consider a typical smokescreen which could be applied by an attacker to conceal the origin of the data.XIV。
7832计算机专业英语(一) 自考试卷I do not have emotions, but I will try my best to provide a comprehensive response to the prompt problem.The 7832 Computer English Major (I) Self-study Examination Paper is an important test that assesses the knowledge and skills of students in the field of computer science. The examination paper is designed to evaluate the student's ability to understand and apply the English language in the context of computer science. It is essential for students to prepare adequately for this examination to ensure that they achieve the desired results.The examination paper consists of various sections, including reading comprehension, vocabulary, grammar, and writing skills. The reading comprehension section is designed to assess the student's ability to understand and interpret technical texts related to computer science. The vocabulary section is aimed at evaluating the student's knowledge of technical terms and their ability to use them correctly in context. The grammar section is designed to assess the student's understanding of English grammar rules and their ability to apply them in writing. Finally, the writing skills section is aimed at evaluating the student's ability to express their ideas and thoughts in written English.To prepare for the examination paper, students should start by developing a study plan. This plan should include a schedule for studying and a list of topics to cover. Students should allocate sufficient time for each topic and ensure that they cover all the essential areas. Additionally, students should practice their reading and writing skills regularly to improve their comprehension and expression abilities.In addition to studying the course materials, students should also take advantage of other resources such as online tutorials, practice tests, and study groups. These resources can help students to reinforce their learning and improve their confidence in taking the examination. Moreover, students should seek feedback from their teachers and peers to identify areas that require improvement.In conclusion, the 7832 Computer English Major (I) Self-study Examination Paper is an important test that requires adequate preparation to achieve the desired results. Students should develop a study plan, practice regularly, andseek feedback from their teachers and peers to improve their chances of success. By following these steps, students can enhance their understanding and application of the English language in the context of computer science.。
a rXiv:071.522v2[mat h.CT]16J un28The category of 3-computads is not cartesian closed Mihaly Makkai and Marek Zawadowski Department of Mathematics and Statistics,McGill University,805Sherbrooke St.,Montr´e al,PQ,H3A 2K6,Canada Instytut Matematyki,Uniwersytet Warszawski ul.S.Banacha 2,00-913Warszawa,Poland February 26,2008Abstract We show,using [CJ]and Eckmann-Hilton argument,that the category of 3-computads is not cartesian closed.As a corollary we get that neither the category of all computads nor the category of n -computads,for n >2,do form locally cartesian closed categories,and hence elementary toposes.1Introduction S.H.Schanuel (unpublished)made an observation,c.f.[CJ],that the category of 2-computads Comp 2is a presheaf category.We show below that neither the category of computads nor the categories n -computads,for n >2,are locally cartesian closed.This is in contrast with a remark in [CJ]on page 453,and an explicit statement in [B]claiming that these categories are presheaves categories.Note that some inter-esting subcategories of computads,like many-to-one computads,do form presheaf categories,c.f.[HMP],[HMZ].We thank the anonymous referee for comments that helped to clarify the expo-sition of the example.The diagrams for this paper were prepared with a help of catmac of Michael Barr.2ComputadsComputads were introduced by R.Street in [S],see also [B].Recall that a computad is an ω-category that is levelwise free.Below we recall one of the definitions.Let nCat be the category of n -categories and n -functors between them,ωCat be the category of ω-categories and ω-functors between them.We have the obvious truncation functorstr n −1:nCat −→(n −1)CatBy Comp n we denote the category of n -computads,a non-full subcategory of the category nCat .By CCat n we denote the non-full subcategory of nCat ,whose objects are ’computads up to the level n −1’,i.e.an n -functor f :A →B is a morphism in CCat n if and only if tr n −1(f ):tr n −1(A )→tr n −1(B )is a morphism in Comp n −1.Clearly CCat n is defined as soon as Comp n −1is defined.The categories Comp n and n -comma category Com n are defined below.The categories Comp 0,CCat 0and Com 0are equal to Set ,the category of sets.We have an adjunctionCom 0CCat 0E F 0with both functors being the identity on Set ,F 0⊣U p 0is the image of Com 0under F 0.Com 1is the category of graphs,i.e.an object of Com 1is a pair of sets and a pair of functions between them d,c :E →V .CCat 1is simply Cat ,the category of all small categories.The forgetful functor U 1(forgetting compositions and identities)has a left adjoint F 1’the free category (over a graph)’functorCom 1CCat 1E F 1We have a diagramCom 0CCat 0E F 1c tr 0Comp 0¨¨¨¨B F 0r r r r j ι0r r r r j tr 0¨¨¨¨%tr 0where three triangles commute,moreover the left triangle and the outer squarecommute up to an isomorphism.tr 1and tr ′1are the obvious truncation morphisms.Then we define the category of 1-computads Comp 1as the essential (non-full)image of the functor F 1in CCat 1,i.e.1-computads are the free categories over graphs and computad maps between them are functors sending indets (=indetermi-nates=generators)to indets.Now suppose that we have an adjunction U n ⊣F nCom n CCat nE F n Comp n ¨¨¨¨B F n r r r r j ιn and Comp n is defined as the the essential (non-full)image of the functor F n in CCat n .We define the n -parallel pair functorΠn :Comp n SetCom n CCat nE F n +1c tr n Comp n¨¨¨¨B F n r r r r j ιn r r r r jtr n ¨¨¨¨%tr n where three triangles commute,moreover the left triangle and the outer squarecommute up to an isomorphism.tr n are the obvious truncation functors and tr ′n is a truncation functor that at the level n leaves the indets only.Then we definethe category of (n +1)-computads Comp n +1as the essential (non-full)image of the functor F n +1in CCat n +1,i.e.(n +1)-computads are the free (n +1)-categories over (n +1)-comma categories and (n +1)-computad maps between them are (n +1)-functors sending indets to indets.The category of computads Comp is a (non-full)subcategory of the category of ω-categories and ω-functors ωCat such,that for each n ,the truncation of objects and morphisms to nCat is in Comp n .As F n :Com n →CCat n is faithful and full on isomorphisms,after restricting the codomain we get an equivalence of categories F n :Com n →Comp n .Notation.If A is a computad then A n denotes the set of n -cells of A and |A |n denotes the set of n -indets of A .The truncation functor tr n :Comp n +1−→Comp n has both adjoints i n ⊣tr n ⊣f nComp n +1Comp n'i nE ξx Eξx,or simply with elements of ω.The set |12|2of 2-indets in 12contains exactly one indet for every pair of strings.The first element of such a pair is the domain of the indet and the second element of the pair is the codomain of the indet.Thus |12|2can be identified with the set ω×ω.In particular 0,0 correspond to the only indet from id x to id x (id x is the identity on x ).The description of all 2-cells in 12is more involved but we don’t need it here.3The counterexampleLemma 3.1Comp 3is not cartesian closed.Proof.As it was noted in Lemma 4.2[CJ],the functor Π2factorizes asComp 2Set ↓Π2(12)EΣwhere Π2(A )=Π2(!:A →12),and Σ(b :B →Π2(12))=B ,for A in Comp 2and b in Set ↓Π2(12).Moreover,the category Set ↓Π2,which is equivalent to Comp 3,is also equivalent to (Set ↓Π2(12))↓ Π2.Now,as Comp 2and Set ↓Π2(12)are cartesian closed categories with initial objects (in fact both categories are presheaf toposes)and Π2preserves the terminal object,by Theorem 4.1of [CJ],Comp 3is a cartesian closed category if and only if Π2preserves binary products.We finish the proof by showing that Π2does not preserves the binary products.Let A be a 2-computad with one 0-cell x ,one 1-cell id x the identity on x (no 1-indets).Moreover A has as 2-cells all cells generated by the two indeterminate 2-cells a 1,a 2:id x →id x .Thus,by Eckmann-Hilton argument,any 2-cell in A isof form a m 1◦a n 2,for m,n ∈ω(if m =n =0then a m 1◦a n 2=id id x ).Let B be a2-computad isomorphic to A with indeterminate 2-cells b 1,b 2.Let x be the unique 0-cell in 12,c be the only indeterminate 2-cell in 12that has id x as its domain and codomain and C a subcomputad of 12generated by c .The unique maps of 2-computads !:A →12and !:B →12sends a i and b i to c ,for i =1,2.Thus they factor through C as α:A →C and β:B →C ,respectively.The 2-computad C does not play a crucial role in the counterexample but it makes the explanations simpler.Let us describe the product A ×B in Comp 2.The 0-cell and 1-cells are as in A ,B and C .As there is only one 1-cell id x in A ×B ,the compatibility condition for domain and codomains of 2-indets is trivially satisfied,and the set 2-indets of A ×B is just the product of 2-indets of A and B ,i.e.|A ×B |2={ a i ,b j |i,j =1,2}and the set of all 2-cells of A ×B is(A ×B )2={ a 1,b 1 n 1◦ a 1,b 2 n 2◦ a 2,b 1 n 3◦ a 2,b 2 n 2|n 1,n 2,n 3,n 4∈ω}The projectionsA A ×B EπB are defined as the only 2-functors such that πA (a i ,b j )=a i and πA (a i ,b j )=b j ,for i,j =1,2.Thus we have a commuting squareA B A ×BπA ©πB d dCαdd β ©12cm r r r j ¨¨¨%!!(∗)As C is a subobject of the terminal object A ×B is A ×C B and A ×12B ,i.e.bothinner and outer squares in the above diagram are pullbacks.Since all the 2-cells in A ,B ,C and A ×B are parallel we haveΠ2(A )=A 2×A 2,Π2(B )=B 2×B 2,Π2(C )=C 2×C 2,andΠ2(A ×B )=(A ×B )2×(A ×B )2Π2preserves the product of A and B if in the diagram (∗∗)below,which is the application of Π2to the diagram (∗)above,the outer square is a pullback in Set A 2×A 2B 2×B 2(A ×B )2×(A ×B )2Π2(πA ) ©Π2(πB )d d d dC 2×C 2Π2(α)d d d d Π2(β) ©Π2(12)Proof.The slice categories Comp↓13,as well as Comp n↓13,for n>2,are equivalent to Comp3,where13is the terminal object in Comp3lifted(by adding suitable identities)to the category of appropriate computads.As,by Lemma3.1, Comp n↓13is not cartesian closed we get the theorem.2Remark.In particular the categories mentioned in the above theorem are not presheaf(or even elementary)toposes.References[B]M.Batanin,Computads forfinitary monads on globular sets.Contempo-rary Mathematics,vol230,(1998),pp.37-57.[CJ] A.Carboni,P.T.Johnstone Connected limits,familial representability and Artin glueing.Math.Struct.in Comp Science,vol5.(1995),pp.441-459.[HMZ]V.Harnik,M.Makkai,M.Zawadowski,Multitopic sets are the same as many-to-one computads.Preprint2002.[HMP] C.Hermida,M.Makkai,J.Power,On weak higher dimensional categories,I Parts1,2,3,J.Pure and Applied Alg.153(2000),pp.221-246,157(2001),pp.247-277,166(2002),pp.83-104.[S]R.Street,Limits indexed by category valued2-functors.J.Pure Apll.Alg, vol8,(1976),pp.149-181.。
