计算机英语(第二版)清华大学出版社课后习题答案

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计算机英语（第二版）清华大学出版社课后习题答案

计算机英语（第二版）清华大学出版社课后习题答案专业英语课后习题答案Chapter1I. 1.Application software2. -128 1273. system software4. hardware system software system5.microcomputers, minicomputers, mainframe computers, supercomputersII. 1. false 2. false 3. false 4. true 5.trueIII. (2)CPU(3)bit(1)integrated circuit(IC)(4) ASCIIIV. (1)编码技术(2)应用软件(3) 浮点数据(4)分时(5)存储容量VIII．1.By using various coding techniques, groups of bits can be made to represent not only binary numbers but also other discrete symbols通过应用多种编码技术，一组二进制数字不但可以表示二进制数据，而且还可以表示其它的离散符号2.System software includes not only the complex programs used by technicians to create application software in the first place but also the organizational programs needed to start up the computer and govern its use of other programs.系统软件不仅包括技术人员用于创建应用软件的复杂程序，而且还包括用于启动计算机和提供给其他程序使用的管理程序。

3.Data are numbers and other binary-code information that are operated on to achieve required computational results.数据是数字和其他的二进制代码信息，通过处理这些数据得到所需要的计算结果。

附A2A3-计算机专业英语(第2版)-邱晓红-清华大学出版社

（1）体现实义性，不拘泥于语法
• 计算机专业英语在应用领域体现了一种实义性，也就是说，计算机专业英语用词节省、句子精练、采用实义词汇而不拘泥于完整语法，只要能说明问题，阐述中心思想即可。例如： ①讲究语法但不拘泥于语法 You will need the user’s manual in order to move up from Sunday driver to UNIX speedster. 为了从UNIX系统的初学者成为一名熟练者，最好使用用户手册。 ②赋予虚词实际意义 There is no if in the case 句中”if”就被赋予了实际意义。句意:这里没有假设的余地。 ③词性转换翻译 His computer is passworded, I can’t use.他给计算机加了密码，我不能使用。 ④典故与专用术语 This is the Achilles’heel which makes MS-DOS an unsuitable for integrating information equipment. 直译：这是阿基理斯的脚跟，它使MS-DOS不适合作为信息设备集成的运载体。参考译文：这是MS-DOS的一个致命弱点，使它不适合作为信息设备集成的媒体。 Achilles’heel 是希腊神话中的一个故事，希神阿基里斯出生后被他的母亲倒提着在冥河水中浸过，除了未浸到水的脚跟外，浑身刀枪不入，所以他的脚跟就成了其致命弱点。
（2）被动语句的翻译
• ①顺序翻译。当原文的被动语句直接翻译成汉语的被动句，又符合汉语习惯时，可顺序翻译。如： It can be automatically generated by a compiler. 可被翻译为：“它可以由（被）一个编译器自动产生”。 ②翻译成主动句。当翻译成汉语时，如不符合中文表达习惯时，可以译为主动句。这类情况比较常见于一些词汇，如expect, require, supply, need, cause等词汇。如：No return value is required.可以翻译为“不需要返回数据”。

计算机英语第二版清华第7单元

Computer Science in EnglishDesigned and Programmed ByJiang TongqiangMiao TianshunTeaching Assisted Courseware1Computer Science in EnglishChapter 7: Applications software2Chapter 7I. Pre-reading Questions1. What is applications software?Reference Key:Application software is the software designed to help you solve problems specific to business or perform specific business tasks. Application software then is the layer of software closest to you. 3Chapter 7I. Pre-reading Questions2. Are there any differences between the systems software and the applications software? Reference Key:(open)4Chapter 7I. Pre-reading Questions3. How many types of applications software do you know?Reference Key:(1) Word processingSpreadsheet applications(3) Personal finance(4) Presentation graphics(5) Database managers(6) Group collaboration(7) Desktop and personal information management(8) Integrated applications(9) Web browsers5过去分词短语designed to help you solve problems…作定语修饰the software。

计算机英语第二版课后习题答案

计算机英语第二版课后习题答案练习答案PART ONE Computer BasicsUnit 1 My ComputerSection A I(Fill in the blanks with the information given inthe text:1( Charles Babbage; Augusta Ada Byron 2( input; output3( VLSI4( workstations; mainframes 5( vacuum; transistors6( instructions; software7( digit; eight; byte8( microminiaturization; chipII(Translate the following terms or phrases from English intoChinese and vice versa:1( artificial intelligence 人工智能2( paper-tape reader 纸空阅读机3( optical computer 光学计算机4( neural network 神经网络5( instruction set 指令集6( parallel processing 平行处理7( difference engine 差分机air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house thefirst element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof andceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters:ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materialsrequire higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,8( versatile logical element 通用逻辑器件9( silicon substrate 硅基10( vacuum tube 真空管(电子管)11( the storage and handling of data 数据的存储与处理12( very large-scale integrated circuit 超大规模集成电路13( central processing unit 中央处理器14( personal computer 个人计算机15( analogue computer 模拟计算机16( digital computer 数字计算机17( general-purpose computer 通用计算机18( processor chip 处理器芯片19( operating instructions 操作指令20( input device 输入设备III(Fill in each of the blanks with one of the words given in the following list, making changes if necessary:We can define a computer as a device that accepts input, processes data, stores data, and produces output. According to the mode of processing, computers are either analog or digital. They can beclassified as mainframes, minicomputers, workstations, or microcomputers. All else (for example, the age of the machine) being equal, this categorization provides some indication of the computer’s speed, size, cost, and abilities.air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia,hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof andceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilationnatural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materialsrequire higher thermal resistance, good thermal insulation properties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,Ever since the advent of computers, there have been constant changes. First-generation computers of historic significance, such as UNIVAC, introduced in the early 1950s, were based on vacuum tubes. Second-generation computers, appearing in the early 1960s, were those in which transistors replaced vacuum tubes. In third-generation computers, dating from the 1960s, integrated circuits replaced transistors. In fourth-generation computers such as microcomputers, which first appeared in the mid-1970s, large-scale integration enabled thousands of circuits to be incorporated on one chip. Fifth-generation computers are expected to combine very-large-scale integration with sophisticated approaches to computing, including artificial intelligence and true distributed processing.IV(Translate the following passage from English into Chinese.A computer system includes a computer, peripheral(外围的)devices, and software. The electric, electronic, and mechanical devices used for processing data are referred to as hardware. Inaddition to the computer itself, the term “hardware” refers to components called peripheral devices air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulationproperties, and thicken the North wall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normallighting,air, keep house air fresh, cooling, moisture, reduce the body temperature of the chicken, which is a measure of the chicken house the first element of the environment. (1) the parameters: ventilation in summer peak demand calculations, 4~5 cubic metres per kilogram of body weight per hour, harmful gas concentration does not exceed 20ppm ammonia, hydrogen sulfide, l0ppm, co 0.15%. (2) ventilation natural ventilation and mechanical ... 5. thermal insulation roof cold season is the region of greatest heat loss, are also hot towel sunlight than any other region, so the roof insulation is the most important region, followed by the walls, if open sheds to make doors and Windows switches freely, and seal well. On most of the walls and roof insulation must be taken or equivalent devices, thermal insulation materials require higher thermal resistance, good thermal insulation properties, and thicken the Northwall must also be taken of thickness and measures such as roof and ceiling canopy. In places with more rain, due the eaves of roofs on both sides extending outward. Pheasants (pheasants) farm building (a) site choose pheasant fields should be selected gaozao terrain, sand texture, well drained, slightly to the South of terrain. Mountain areas should be selected under the sunny, spacious, ventilation, sunshine, where the drainage is good. Pheasant farms should be built in a quiet, safe place, away from residential areas, factories, the main traffic artery, but taking into account feed, transportation problems. A clean water source, water is not contaminated. To have a reliable power supply, not only to maintain normal lighting,that expand the computer’s input, output, and storage capabilities. Computer hardware in and of itself does not provide a particularly useful mind tool. To be useful, a computer requires a set of instructions, called software or a computer program, which tells the computer how to perform a particular task. Computers become even more effective when connected to other computers in a network so users can share information.计算机系统包括计算机、外围设备和软件。

计算机英语第二版译文和答案(1—4单元及第9单元)

《计算机英语》参考译文和练习答案目录第一单元 (2)课文A：计算机概览 (2)第二单元 (4)课文A：计算机硬件 (4)第三单元 (7)课文A：操作系统 (7)第四单元 (10)课文A：编程语言 (10)第九单元 (12)课文A：计算机网络 (12)第一单元课文A：计算机概览一、引言计算机是一种电子设备，它能接收一套指令或一个程序，然后通过对数值数据进行运算或者对其他形式的信息进行处理来执行该程序。

如果没有计算机的发展，现代的高科技世界是不可能产生的。

在整个社会，不同型号和不同大小的计算机被用于存储和处理各种数据，从政府保密文件、银行交易到私人家庭账目。

计算机通过自动化技术开辟了制造业的新纪元，而且它们也增强了现代通信系统的性能。

在几乎每一个研究和应用技术领域，从构建宇宙模型到产生明天的气象报告，计算机无不是必要的工具，并且它们的应用本身就开辟了人们推测的新领域。

数据库服务和计算机网络使各种各样的信息源可供使用。

同样的先进技术也使侵犯个人隐私和商业秘密成为可能。

计算机犯罪已经成为作为现代技术代价组成部分的许多风险之一。

二、历史第一台加法机，数字计算机的先驱，是1642年由法国科学家、数学家兼哲学家布莱斯•帕斯卡设计的。

这个装置使用了一系列有10个齿的轮子，每个齿代表从0到9的一个数字。

轮子互相连接，从而通过按照正确的齿数向前移动轮子，就可以将数字彼此相加。

在17世纪70年代，德国哲学家兼数学家戈特弗里德•威廉•莱布尼兹对这台机器进行了改良，设计了一台也能做乘法的机器。

法国发明家约瑟夫―玛丽•雅卡尔，在设计自动织机时，使用了穿孔的薄木板来控制复杂图案的编织。

在19世纪80年代期间，美国统计学家赫尔曼•何勒里斯，想出了使用类似雅卡尔的木板那样的穿孔卡片来处理数据的主义。

通过使用一种将穿孔卡片从电触点上移过的系统，他得以为1890年的美国人口普查汇编统计信息。

1、分析机也是在19世纪，英国数学家兼发明家查尔斯•巴比奇，提出了现代数字计算机的原理。

计算机专业英语(第二版)参考答案

写在前面的话各位老师：您们好。

欢迎使用劳动版专业英语教材。

如果您在使用中发现问题，或者有任何建议和意见，可与我们联系：联系人：赵硕联系电话：64962011电子邮箱：zhaos@计算机专业英语（第二版）参考答案Unit 1Lesson1Look and LearnA 1 Monitor 显示器 2 Modem 调制解调器 3 System Unit 系统单元，主机4 Mouse 鼠标 5 Speaker音箱 6 Printer 打印机7 Keyboard 键盘B 1.麦克风可以用来刻录声音。

2.操纵杆是为玩游戏而设计的。

3.多数的扫描仪可以既扫描图片又扫描文本。

4.绘图板被用于图形设计。

DialogueA 略B 1. Mary is good at computer.2. Because it hasn’t the computer software system, Mike’s computer can’t work.3. Mike is a computer outsider.Reading179A Input Device: keyboard, mouse, microphone, scanner, graphics tabletOutput Device: monitor, speaker, printerB Input device (输入设备) convert into(把….转变为) output device(输出设备) translate into (翻译成)Lesson 2Look and Learn1 BIOS-ROM 芯片2 Flash memory 闪存3 memory bank内存条4 CMOS setup CMOS 设置DialogueA Install configuration error utility invalidB1.Mary works at PC Support Center.2.Bill should run the CMOS setup utility to solve the problem.ReadingA 1 b 2 a 3 d 4 c 5 eB 1. Random access memory (RAM); RAM is volatile storage because everything in most types of RAM is lost as soon as the computer is turned off.2. Cache memory; Cache memory improves processing by acting as a temporary high-speed holding area between the memory and the CPU.3. Flash RAM; Flash RAM can retain data even if power is disrupted.1804. Read only memory (ROM); ROM chips are not volatile and cannot be changed by the user.1 Floppy disk 软盘2 CD-ROM 光驱3 Hard disk 硬盘4 Flash Disk 闪存盘；u盘 5 mobile hard disk 移动硬盘DialogueB１. Yes 2. Yes 3. No 4. YesReadingA 1 A 2 Bseries prevent unrecoverable comprise dustB concentric181Lesson 4Look and Learn1. CPU slot (CPU插槽)2. Memory slots （内存插槽）3.AGP slot（AGP 插槽）4.PCI slots（PCI插槽）5. Power connector （电源接口）6.Input/Output ports（输入输出端口）DialogueA interface socket battery slot processorB 1.Because the battery of the motherboard is getting low.2. The socket types of both the processor and the motherboard have to be the same for them to work with each other. The motherboard must have updated BIOS in order for the certain CPUs to work right.ReadingA tower medium component expansion serial specializeB 1 No 2 Yes 3 NoUnit 2Lesson1Look and Learn1. desktop2. Pop-up Menu3. Folder4. Screen saverDialogue182B1.His card is an external sound card.2.Windows will look for and install a driver automatically.ReadingA 1. b 2. d 3. c 4. aB1. Operating system recognizes input from the keyboard and sends output to the display screen.2. Operating system keeps track of files and directories on the disk.3. Operating system controls peripheral devices such as disk drives and printers.Lesson2Look and LearnA Word Access Excel PowerPointB 1.Title bar 2.Menu bar 3. Toolbar 4.Row 5. Column 6. Cell 7.Statusbar DialogueA1. word2. icon3. dialog boxB1. NO2. Yes3. YesReadingA command grid software interface interact formulaB 1. a 2. b 3. cLesson3Look and LearnA. PhotoshopB. FlashC. 3DS MAXD. After EffectsDialogue1. web pages2. Flah Player3. Plug-in4. browser183ReadingA visual audio technology interactivity navigationalB 1. B 2.C 3. BLesson4Look and Learn1. carbon copy2. subject3. attachment4.salutationDialogueA 1. Yes 2. NO 3. Yes 4. NoB 1. set up 2. log 3. Outlook ExpressReadingA filter intention instruction confidentialB 1. D 2.C 3. A 4. BUnit 3Lesson1Look and Learnwork interface card2.Hub3. Switch4.Wireless Router5.CableDialogueB 1. NO 2. Yes 3. Yes 4. NOReadingA 1. b 2. d 3. a 4. cB 1. a 2.c 3. bLesson2Look and Learn1841. Domain name2. Browser3. Website4. ProtocolDialogueA 1.ADSL 2. ISP 3. IP and DNS addressReadingA cable upload motorway permanentB 1. c 2. a 3. d 4. bLesson3Look and Learn1. bus2. star3. ring4. treeDialogueA 1.Toplogy 2. star 3. dynamicB 1. Yes 2. NO 3. Yes 4. NOReadingA format device protocol destinationB 1. c 2. a 3. bLesson4Look and Learn1. firewall2. antivirus3. spam4. virusDialogueA 1.Control panel 2. antivirus 3. securityB 1. NO 2. Yes 3. NO 4. YesReadingA hack management maintenance attackB 1. a 2. b185Unit 4Lesson1Look and learn（1）销售副总（2）营销经理（3）销售代表（4）销售助理（5）采买部经理（6）采购员DialogueAB 1. ABC Company and Huaxia Commercial Company2.products design3. configuration and priceReadingA 1.No 2. Yes 3. Yes 4. YesB 1.B 2.B 3.DLesson 2Look and learn1. discount stores2. catalogue3. promotion4. market share Dialogue186187B1. Not at all.a.你怎么啦？有什么问题吗？ 2. What ’s wrong with you?b.期待再次相见。

计算机英语第二版清华第2单元

Computer Science in EnglishDesigned and Programmed ByJiang TongqiangMiao TianshunTeaching Assisted Courseware1Computer Science in EnglishChapter 2:Computer System Organization2Chapter 2I. Pre-reading Questions1. What is a computer system?Reference Key:A computer system consists of hardware system and software system. The hardware of the computer is usually divided into three major parts or three primary subsystems: the CPU, the memory subsystem, and the I/O subsystem. The software system can be divided into two broad categories: application software and system software..3Chapter 2I. Pre-reading Questions2. How many levels of memory hierarchy do you know? Name at least three of them. Reference Key:The memory hierarchy system consists of three levels.They are a cache ,the main memory and the auxiliary memory.4Chapter 2I. Pre-reading Questions3. Identify the types of data transfer modes.Reference Key:Data transfer has three modes: programmed I/O, interrupts, and direct memory access (DMA).5being executed：为分词短评作定语，修饰program, 意思是“由……执行”。

计算机英语学生用书(第二版)习题答案chapter 05.

IV. Exercises Comprehension I. Filling Blanks
II. True or False III.Answering Questions IV. Translating Terms
Vocabulary
V. Selecting Words
Collocation Summary
• First generation—— machine language • Second generation—— assembly language • Third generation—— high-level languages • Fourth generation—— very-high-level
5. T
IIV. Exercises—Comprehension
III. Answer the following question.
1. List the five steps of programming.
1. Programming is traditionally a five-step problem-solving process. The five steps are as follows: The first step: Problem definition The second step: Program design The third step: Program coding The fourth step: Program testing The fifth step: Program documentation and maintenance
languages • Fifth generation—— natural languages

计算机专业英语(第二版)参考答案

写在前面的话各位老师：您们好。

欢迎使用劳动版专业英语教材。

2.操纵杆是为玩游戏而设计的。

3.多数的扫描仪可以既扫描图片又扫描文本。

4.绘图板被用于图形设计。

算法分析与设计第二版英文版 (潘彦著) 清华大学出版社课后答案--solu9

This ﬁle contains the exercises,hints,and solutions for Chapter 9of the book ”Introduction to the Design and Analysis of Algorithms,”2nd edition,byA.Levitin.The problems that might be challenging for at least some students are marked by ;those that might be diﬃcult for a majority of students are marked by .Exercises 9.11.Give an instance of the change-making problem for which the greedy al-gorithm does not yield an optimal solution.2.Write a pseudocode of the greedy algorithm for the change-making prob-lem,with an amount n and coin denominations d 1>d 2>...>d m as its input.What is the time eﬃciency class of your algorithm?3.Consider the problem of scheduling n jobs of known durations t 1,...,t n for execution by a single processor.The jobs can be executed in any order,one job at a time.You want to ﬁnd a schedule that minimizes the total time spent by all the jobs in the system.(The time spent by one job in the system is the sum of the time spent by this job in waiting plus the time spent on its execution.)Design a greedy algorithm for this problem. Does the greedy algo-rithm always yield an optimal solution?4.Design a greedy algorithm for the assignment problem (see Section 3.4).Does your greedy algorithm always yield an optimal solution?5.Bridge crossing revisited Consider the generalization of the bridge cross-ing puzzle (Problem 2in Exercises 1.2)in which we have n >1people whose bridge crossing times are t 1,t 2,...,t n .All the other conditions of the problem remain the same:at most two people at the time can cross the bridge (and they move with the speed of the slower of the two)and they must carry with them the only ﬂashlight the group has.Design a greedy algorithm for this problem and ﬁnd how long it willtake to cross the bridge by using this algorithm.Does your algorithm yields a minimum crossing time for every instance of the problem?If it does–prove it,if it does not–ﬁnd an instance with the smallest number of people for which this happens.6.Bachet-Fibonacci weighing problem Find an optimal set of n weights {w 1,w 2,...,w n }so that it would be possible to weigh on a balance scale any integer load in the largest possible range from 1to W ,provided a. weights can be put only on the free cup of the scale.b. weights can be put on both cups of the scale.1课后答案网 w w w .k h d a w .c o m7.a.Apply Prim’s algorithm to the following graph.Include in the priority queue all the vertices not already in the tree.b.Apply Prim’s algorithm to the following graph.Include in the priority queue only the fringe vertices (the vertices not in the current tree which are adjacent to at least one tree vertex).8.The notion of a minimum spanning tree is applicable to a connected weighted graph.Do we have to check a graph’s connectivity before ap-plying Prim’s algorithm or can the algorithm do it by itself?9.a.How can we use Prim’s algorithm to ﬁnd a spanning tree of a connected graph with no weights on its edges?b.Is it a good algorithm for this problem?10. Prove that any weighted connected graph with distinct weights hasexactly one minimum spanning tree.11.Outline an eﬃcient algorithm for changing an element’s value in a min-heap.What is the time eﬃciency of your algorithm?2课后答案网 w h d a w .c o mHints to Exercises 9.11.As coin denominations for your counterexample,you may use,among a multitude of other possibilities,the ones mentioned in the text:d 1=7,d 2=5,d 3=1.2.You may use integer divisions in your algorithm.3.Considering the case of two jobs might help.Of course,after forming a hypothesis,you will have to either prove the algorithm’s optimality for an arbitrary input or ﬁnd a speciﬁc counterexample showing that it is not the case.4.You can apply the greedy approach either to the entire cost matrix or to each of its rows (or columns).5.Simply apply the greedy approach to the situation at hand.You may assume that t 1≤t 2≤...≤t n .6.For both versions of the problem,it is not diﬃcult to get to a hypothesis about the solution’s form after considering the cases of n =1,2,and 3.It is proving the solutions’optimality that is at the heart of this problem.7.a.Trace the algorithm for the graph given.An example can be found in the text of the section.b.After the next fringe vertex is added to the tree,add all the unseen vertices adjacent to it to the priority queue of fringe vertices.8.Applying Prim’s algorithm to a weighted graph that is not connected should help in answering this question.9.a.Since Prim’s algorithm needs weights on a graph’s edges,some weights have to be assigned.b.Do you know other algorithms that can solve this problem?10.Strictly speaking,the wording of the question asks you to prove two things:the fact that at least one minimum spanning tree exists for any weighted connected graph and the fact that a minimum spanning tree is unique if all the weights are distinct numbers.The proof of the former stems from the obvious observation about ﬁniteness of the number of spanning trees for a weighted connected graph.The proof of the latter can be obtained by repeating the correctness proof of Prim’s algorithm with a minor adjustment at the end.11.Consider two cases:the key’s value was decreased (this is the case needed for Prim’s algorithm)and the key’s value was increased.3课后答案网 w w w .k h d a w .c o mSolutions to Exercises 9.11.Here is one of many such instances:For the coin denominations d 1=7,d 2=5,d 3=1and the amount n =10,the greedy algorithm yields one coin of denomination 7and three coins of denomination 1.The actual optimal solution is two coins of denomination 5.2.Algorithm Change (n,D [1..m ])//Implements the greedy algorithm for the change-making problem //Input:A nonnegative integer amount n and//a decreasing array of coin denominations D//Output:Array C [1..m ]of the number of coins of each denomination //in the change or the ”no solution”messagefor i ←1to m doC [i ]← n/D [i ]n ←n mod D [i ]if n =0return Celse return ”no solution”The algorithm’s time eﬃciency is in Θ(m ).(We assume that integer di-visions take a constant time no matter how big dividends are.)Note also that if we stop the algorithm as soon as the remaining amount becomes 0,the time eﬃciency will be in O (m ).3.a.Sort the jobs in nondecreasing order of their execution times and exe-cute them in that order.b.Yes,this greedy algorithm always yields an optimal solution.Indeed,for any ordering (i.e.,permutation)of the jobs i 1,i 2,...,i n ,the total time in the system is given by the formula t i 1+(t i 1+t i 2)+...+(t i 1+t i 2+...+t i n )=nt i 1+(n −1)t i 2+...+t i n .Thus,we have a sum of numbers n,n −1,...,1multiplied by “weights”t 1,t 2,...t n assigned to the numbers in some order.To minimize such a sum,we have to assign smaller t ’s to larger numbers.In other words,the jobs should be executed in nondecreasing order of their execution times.Here is a more formal proof of this fact.We will show that if jobs are ex-ecuted in some order i 1,i 2,...,i n ,in which t i k >t i k +1for some k,then the total time in the system for such an ordering can be decreased.(Hence,no such ordering can be an optimal solution.)Let us consider the other job ordering,which is obtained by swapping the jobs k and k +1.Obvi-ously,the time in the systems will remain the same for all but these two 4课后答案网 w w w .k h d a w .c o mjobs.Therefore,the diﬀerence between the total time in the system for the new ordering and the one before the swap will be[(k −1j =1t i j +t i k +1)+(k −1j =1t i j +t i k +1+t i k )]−[(k −1j =1t i j +t i k )+(k −1j =1t i j +t i k +t i k +1)]=t i k +1−t i k <0.4.a.The all-matrix version:Repeat the following operation n times.Select the smallest element in the unmarked rows and columns of the cost matrix and then mark its row and column.The row-by-row version:Starting with the ﬁrst row and ending with the last row of the cost matrix,select the smallest element in that row which is not in a previously marked column.After such an element is selected,mark its column to prevent selecting another element from the same col-umn.b.Neither of the versions always yields an optimal solution.Here isa simple counterexample:C = 122100 5.Repeat the following step n −2times:Send to the other side the pair of two fastest remaining persons and then return the ﬂashlight with the fastest person.Finally,send the remaining two people together.Assuming that t 1≤t 2≤...≤t n ,the total crossing time will be equal to (t 2+t 1)+(t 3+t 1)+...+(t n −1+t 1)+t n =ni =2t i +(n −2)t 1=n i =1t i +(n −3)t 1.Note:For an algorithm that always yields a minimal crossing time,seeGünter Rote,“Crossing the Bridge at Night,”EATCS Bulletin,vol.78(October 2002),241—246.The solution to the instance of Problem 2in Exercises 1.2shows that the greedy algorithm doesn’t always yield the minimal crossing time for n >3.No smaller counterexample can be given as a simple exhaustive check for n =3demonstrates.(The obvious solution for n =2is the one generated by the greedy algorithm as well.)5课后答案网 w w w .kh d a w .c o m6.a.Let’s apply the greedy approach to the ﬁrst few instances of the problem in question.For n =1,we have to use w 1=1to balance weight 1.For n =2,we simply add w 2=2to balance the ﬁrst previously unattainable weight of 2.The weights {1,2}can balance every integral weights up to their sum 3.For n =3,in the spirit of greedy thinking,we take the next previously unattainable weight:w 3=4.The three weights {1,2,4}allow to weigh any integral load l between 1and their sum 7,with l ’s binary expansion indicating the weights needed for load l :Generalizing these observations,we should hypothesize that for any posi-tive integer n the set of consecutive powers of 2{w i =2i −1,i =1,2,...n }makes it possible to balance every integral load in the largest possible range,which is up to and including n i =12i −1=2n −1.The fact that every integral weight l in the range 1≤l ≤2n −1can be balanced with this set of weights follows immediately from the binary expansion of l,which yields the weights needed for weighing l.(Note that we can obtain the weights needed for a given load l by applying to it the greedy algorithm for the change-making problem with denominations d i =2i −1,i =1,2,...n.)In order to prove that no set of n weights can cover a larger range of consecutive integral loads,it will suﬃce to note that there are just 2n −1nonempty selections of n weights and,hence,no more than 2n −1sums they yield.Therefore,the largest range of consecutive integral loads they can cover cannot exceed 2n −1.[Alternatively,to prove that no set of n weights can cover a larger range of consecutive integral loads,we can prove by induction on i that if any mul-tiset of n weights {w i ,i =1,...,n }–which we can assume without loss of generality to be sorted in nondecreasing order–can balance every integral load starting with 1,then w i ≤2i −1for i =1,2,...,n.The basis checks out immediately:w 1must be 1,which is equal to 21−1.For the general case,assume that w k ≤2k −1for every 1≤k <i.The largest weight the ﬁrst i −1weights can balance is i −1k =1w k ≤ i −1k =12k −1=2i −1−1.If w i were larger than 2i ,then this load could have been balanced neither with the ﬁrst i −1weights (which are too light even taken together)nor with the weights w i ≤...≤w n (which are heavier than 2i even individ-ually).Hence,w i ≤2i −1,which completes the proof by induction.This immediately implies that no n weights can balance every integral load up to the upper limit larger than n i =1w i ≤ n i =12i −1=2n −1,the limit attainable with the consecutive powers of 2weights.]b.If weights can be put on both cups of the scale,then a larger range can 6课后答案网 w w w .k h d a w .be reached with n weights for n >1.(For n =1,the single weight still needs to be 1,of course.)The weights {1,3}enable weighing of every integral load up to 4;the weights {1,3,9}enable weighing of every inte-gral load up to 13,and,in general,the weights {w i =3i −1,i =1,2,...,n }enable weighing of every integral load up to and including their sum of n i =13i −1=(3n −1)/2.A load’s expansion in the ternary system indicates the weights needed.If the ternary expansion contains only 0’s and 1’s,the load requires putting the weights corresponding to the 1’s on the opposite cup of the balance.If the ternary expansion of load l,l ≤(3n −1)/2,contains one or more 2’s,we can replace each 2by (3-1)to represent it in the form l =n i =1βi 3i −1,where βi ∈{0,1,−1},n = log 3(l +1) .In fact,every positive integer can be uniquely represented in this form,obtained from its ternary expansion as described above.For example,5=123=1·31+2·30=1·31+(3−1)·30=2·31−1·30=(3−1)·31−1·30=1·32−1·31−1·30.(Note that if we start with the rightmost 2,after a simpliﬁcation,the new rightmost 2,if any,will be at some position to the left of the starting one.This proves that after a ﬁnite number of such replacements,we will be able to eliminate all the 2’s.)Using the representation l = n i =1βi 3i −1,we can weigh load l by placing all the weights w i =3i −1for negative βi ’s along with the load on one cup of the scale and all the weights w i =3i −1for positive βi ’s on the opposite cup.Now we’ll prove that no set of n weights can cover a larger range of con-secutive integral loads than (3n −1)/2.Each of the n weights can be either put on the left cup of the scale,or put on the right cup,or not to be used at all.Hence,there are 3n −1possible arrangements of the weights on the scale,with each of them having its mirror image (where all the weights are switched to the opposite pan of the scale).Eliminating this symmetry,leaves us withjust (3n −1)/2arrangements,which can weight at most (3n −1)/2diﬀerent integral loads.Therefore,the largest range of consecutive integral loads they can cover cannot exceed (3n −1)/2.7.a.Apply Prim’s algorithm to the following graph:7课后答案网 w w w.k h d a w .c o mthe edges ae,eb,ec,and cd.b.Apply Prim’s algorithm to the following graph:the edges ab,be,ed,dc,ef,ei,ij,cg,gh,il,gk.8.There is no need to check the graph’s connectivity because Prim’s algo-rithm can do it itself.If the algorithm reaches all the graph’s vertices (via edges of ﬁnite lengths),the graph is connected,otherwise,it is not.9.a.The simplest and most logical solution is to assign all the edge weights to 1.8课a w .c o mb.Applying a depth-ﬁrst search (or breadth-ﬁrst search)traversal to get a depth-ﬁrst search tree (or a breadth-ﬁrst search tree),is conceptually simpler and for sparse graphs represented by their adjacency lists faster.10.The number of spanning trees for any weighted connected graph is a pos-itive ﬁnite number.(At least one spanning tree exists,e.g.,the one obtained by a depth-ﬁrst search traversal of the graph.And the number of spanning trees must be ﬁnite because any such tree comprises a subset of edges of the ﬁnite set of edges of the given graph.)Hence,one can always ﬁnd a spanning tree with the smallest total weight among the ﬁnite number of the candidates.Let’s prove now that the minimum spanning tree is unique if all the weights are distinct.We’ll do this by contradiction,i.e.,by assuming that there exists a graph G =(V,E )with all distinct weights but with more than one minimum spanning tree.Let e 1,...,e |V |−1be the list of edges com-posing the minimum spanning tree T P obtained by Prim’s algorithm with some speciﬁc vertex as the algorithm’s starting point and let T be an-other minimum spanning tree.Let e i =(v,u )be the ﬁrst edge in the list e 1,...,e |V |−1of the edges of T P which is not in T (if T P =T ,such edge must exist)and let (v,u )be an edge of T connecting v with a vertex not in the subtree T i −1formed by {e 1,...,e i −1}(if i =1,T i −1consists of vertex v only).Similarly to the proof of Prim’s algorithms correctness,let us replace (v,u )by e i =(v,u )in T .It will create another spanning tree,whose weight is smaller than the weight of T because the weight of e i =(v,u )is smaller than the weight of (v,u ).(Since e i was chosen by Prim’s algorithm,its weight is the smallest among all the weights on the edges connecting the tree vertices of the subtree T i −1and the vertices adjacent to it.And since all the weights are distinct,the weight of (v,u )must be strictly greater than the weight of e i =(v,u ).)This contradicts the assumption that T was a minimum spanning tree.11.If a key’s value in a min-heap was decreased,it may need to be pushedup (via swaps)along the chain of its ancestors until it is smaller than or equal to its parent or reaches the root.If a key’s value in a min-heap was increased,it may need to be pushed down by swaps with the smaller of its current children until it is smaller than or equal to its children or reaches a leaf.Since the height of a min-heap with n nodes is equal to log 2n (by the same reason the height of a max-heap is given by this formula–see Section 6.4),the operation’s eﬃciency is in O (log n ).(Note:The old value of the key in question need not be known,of paring the new value with that of the parent and,if the min-heap condition holds,with the smaller of the two children,will suﬃce.)9课后答案网 w w w.k h d a w .c o mExercises 9.21.Apply Kruskal’s algorithm to ﬁnd a minimum spanning tree of the follow-ing graphs.a.b.2.Indicate whether the following statements are true or false:a.If e is a minimum-weight edge in a connected weighted graph,it must be among edges of at least one minimum spanning tree of the graph.b.If e is a minimum-weight edge in a connected weighted graph,it must be among edges of each minimum spanning tree of the graph.c.If edge weights of a connected weighted graph are all distinct,the graph must have exactly one minimum spanning tree.d.If edge weights of a connected weighted graph are not all distinct,the graph must have more than one minimum spanning tree.3.What changes,if any,need to be made in algorithm Kruskal to make it ﬁnd a minimum spanning forest for an arbitrary graph?(A minimum spanning forest is a forest whose trees are minimum spanning trees of the graph’s connected components.)10课后答案网h d a w .c o m4.Will either Kruskal’s or Prim’s algorithm work correctly on graphs that have negative edge weights?5.Design an algorithm for ﬁnding a maximum spanning tree –a spanning tree with the largest possible edge weight–of a weighted connected graph.6.Rewrite the pseudocode of Kruskal’s algorithm in terms of the operations of the disjoint subsets’ADT.7. Prove the correctness of Kruskal’s algorithm.8.Prove that the time eﬃciency of ﬁnd (x )is in O (log n )for the union-by-size version of quick union.9.Find at least two Web sites with animations of Kruskal’s and Prim’s al-gorithms.Discuss their merits and demerits..10.Design and conduct an experiment to empirically compare the eﬃcienciesof Prim’s and Kruskal’s algorithms on random graphs of diﬀerent sizes and densities.11. Steiner tree Four villages are located at the vertices of a unit squarein the Euclidean plane.You are asked to connect them by the shortest network of roads so that there is a path between every pair of the villages along those roads.Find such a network.11课后答案网ww w.kh d aw .c omHints to Exercises 9.21.Trace the algorithm for the given graphs the same way it is done for another input in the section.2.Two of the four assertions are true,the other two are false.3.Applying Kruskal’s algorithm to a disconnected graph should help to an-swer the question.4.The answer is the same for both algorithms.If you believe that the algorithms work correctly on graphs with negative weights,prove this assertion;it you believe this is not to be the case,give a counterexample for each algorithm.5.Is the general trick of transforming maximization problems to their mini-mization counterparts (see Section6.6)applicable here?6.Substitute the three operations of the disjoint subsets’ADT–makeset (x ),ﬁnd (x ),and union (x,y )–in the appropriate places of the pseudocode given in the section.7.Follow the plan used in Section 9.1to prove the correctness of Prim’s algorithm.8.The argument is very similar to the one made in the section for the union-by-size version of quick ﬁnd.9.You may want to take advantage of the list of desirable characteristics in algorithm visualizations,which is given in Section 2.7.10.n/a11.The question is not trivial because introducing extra points (called Steinerpoints )may make the total length of the network smaller than that of a minimum spanning tree of the square.Solving ﬁrst the problem for three equidistant points might give you an indication how a solution to the problem in question could look like.12课后答案网ww w.kh d aw .c omSolutions to Exercises9.21.a.后课13b.⇒⇒⇒⇒⇒⇒14课c⇒⇒⇒⇒⇒课2.a.True.(Otherwise,Kruskal’s algorithm would be invalid.)b.False.As a simple counterexample,consider a complete graph withthree vertices and the same weight on its three edgesc.True(Problem10in Exercises9.1).15d.False (see,for example,the graph of Problem 1a).3.Since the number of edges in a minimum spanning forest of a graph with |V |vertices and |C |connected components is equal to |V |−|C |(this for-mula is a simple generalization of |E |=|V |−1for connected graphs),Kruskal (G )will never get to |V |−1tree edges unless the graph is con-nected.A simple remedy is to replace the loop while ecounter <|V |−1with while k <|E |to make the algorithm stop after exhausting the sorted list of its edges.4.Both algorithms work correctly for graphs with negative edge weights.One way of showing this is to add to all the weights of a graph with negative weights some large positive number.This makes all the new weights positive,and one can “translate”the algorithms’actions on the new graph to the corresponding actions on the old one.Alternatively,you can check that the proofs justifying the algorithms’correctness do not depend on the edge weights being nonnegative.5.Replace each weight w (u,v )by −w (u,v )and apply any minimum spanning tree algorithm that works on graphs with arbitrary weights (e.g.,Prim’s or Kruskal’s algorithm)to the graph with the new weights.6.Algorithm Kruskal (G )//Kruskal’s algorithm with explicit disjoint-subsets operations //Input:A weighted connected graph G = V,E//Output:E T ,the set of edges composing a minimum spanning tree of G sort E in nondecreasing order of the edge weights w (e i 1)≤...≤w (e i |E |)for each vertex v ∈V make (v )E T ←∅;ecounter ←0//initialize the set of tree edges and its size k ←0//the number of processed edges while ecounter <|V |−1k ←k +1if ﬁnd (u )=ﬁnd (v )//u,v are the endpoints of edge e i kE T ←E T ∪{e i k };ecounter ←ecounter +1union (u,v )return E T 7.Let us prove by induction that each of the forests F i ,i =0,...,|V |−1,of Kruskal’s algorithm is a part (i.e.,a subgraph)of some minimum span-ning tree.(This immediately implies,of course,that the last forest in the sequence,F |V |−1,is a minimum spanning tree itself.Indeed,it contains all vertices of the graph,and it is connected because it is both acyclic and has |V |−1edges.)The basis of the induction is trivial,since F 0is16课后答案网ww w.kh d aw .c ommade up of |V |single-vertex trees and therefore must be a subgraph of any spanning tree of the graph.For the inductive step,let us assume that F i −1is a subgraph of some minimum spanning tree T .We need to prove that F i ,generated from F i −1by Kruskal’s algorithm,is also a part of a minimum spanning tree.We prove this by contradiction by assuming that no minimum spanning tree of the graph can contain F i .Let e i =(v,u )be the minimum weight edge added by Kruskal’s algorithm to forest F i −1to obtain forest F i .(Note that vertices v and u must belong to diﬀerent trees of F i −1–otherwise,edge (v,u )would’ve created a cycle.)By our assumption,e i cannot belong to T .Therefore,if we add e i to T ,a cycle must be formed (see the ﬁgure below).In addition to edge e i =(v,u ),this cycle must contain another edge (v ,u )connecting a vertex v in the same tree of F i −1as v to a vertex u not in that tree.(It is possible that v coincides with v or u coincides with u but not both.)If we now delete the edge (v ,u )from this cycle,we will obtain another spanning tree of the entire graph whose weight is less than or equal to the weight of T since the weight of e i is less than or equal to the weight of (v ,u ).Hence,this spanning tree is a minimum spanning tree,which contradicts the assumption that no minimum spanning tree contains F i .This com-pletes the correctness proof of Kruskal’s algorithm.8.In the union-by-size version of quick-union ,each vertex starts at depth 0of its own tree.The depth of a vertex increases by 1when the tree it is in is attached to a tree with at least as many nodes during a union operation.Since the total number of nodes in the new tree containing the node is at least twice as much as in the old one,the number of such increases cannot exceed log 2n.Therefore the height of any tree (which is the largest depth of the tree’s nodes)generated by a legitimate sequence of unions will not exceed log 2n.Hence,the eﬃciency of ﬁnd (x )is in O (log n )because ﬁnd (x )traverses the pointer chain from the x ’s node to the tree’s root.9.n/a10.n/a17课后答案.kh d aw .c om11.The minimum Steiner tree that solves the problem is shown below.(Theother solution can be obtained by rotating the ﬁgure 90◦.)A popular discussion of Steiner trees can be found in “Last Recreations:Hydras,Eggs,and Other Mathematical Mystiﬁcations”by Martin Gard-ner.In general,no polynomial time algorithm is known for ﬁnding a minimum Steiner tree;moreover,the problem is known to be NP -hard (see Section 11.3).For the state-of-the-art information,see,e.g.,The Steiner Tree Page at /steiner/.18课后答案网ww w.kc omExercises 9.31.Explain what adjustments if any need to be made in Dijkstra’s algorithm and/or in an underlying graph to solve the following problems.a.Solve the single-source shortest-paths problem for directed weighted graphs.b.Find a shortest path between two given vertices of a weighted graph or digraph.(This variation is called the single-pair shortest-path prob-lem .)c.Find the shortest paths to a given vertex from each other vertex of a weighted graph or digraph.(This variation is called the single-destination shortest-paths problem .)d.Solve the single-source shortest-path problem in a graph with nonneg-ative numbers assigned to its vertices (and the length of a path deﬁned as the sum of the vertex numbers on the path).2.Solve the following instances of the single-source shortest-paths problem with vertex a as the source:a.b.3.Give a counterexample that shows that Dijkstra’s algorithm may not work for a weighted connected graph with negative weights.19课案w w.kh d aw .c om4.Let T be a tree constructed by Dijkstra’s algorithm in the process of solving the single-source shortest-path problem for a weighted connected graph G .a.True or false:T is a spanning tree of G ?b.True or false:T is a minimum spanning tree of G ?5.Write a pseudocode of a simpler version of Dijkstra’s algorithm that ﬁnds only the distances (i.e.,the lengths of shortest paths but not shortest paths themselves)from a given vertex to all other vertices of a graph represented by its weight matrix.6. Prove the correctness of Dijkstra’s algorithm for graphs with positive weights.7.Design a linear-time algorithm for solving the single-source shortest-paths problem for dags (directed acyclic graphs)represented by their adjacency lists.8.Design an eﬃcient algorithm for ﬁnding the length of a longest path in a dag.(This problem is important because it determines a lower bound on the total time needed for completing a project composed of precedence-constrained tasks.)9.Shortest-path modeling Assume you have a model of a weighted con-nected graph made of balls (representing the vertices)connected by strings of appropriate lengths (representing the edges).a.Describe how you can solve the single-pair shortest-path problem with this model .b.Describe how you can solve the single-source shortest-paths problem with this model .10.Revisit Problem 6in Exercises 1.3about determining the best route fora subway passenger to take from one designated station to another in a well-developed subway system like those in Washington,DC and London,UK.Write a program for this task.20课后答案网ww w.kh d aw .c om。

计算机英语第二版_清华大学_姜同强主编翻译

第1 章计算机系统概述数字计算机也称为电子计算机或计算机．它无处不在．很难发现有哪个领域没有使用计算机．在本章中．我们将介绍数字计算机、数据类型、计算机的发展和计算机分类，1 . 1 数字计算机数字计算机是处理各种计算任务的数字系统。

“数字”的意思是计算机中的信息可由具有部分离散值的变量来表示，这些离散值的内部处理是由具有部分离散状态的部件完成的．例如．十进制数字O , l , 2 ，… 9 ．即提供了10 个离散值。

第一台电子数字计算机产生于20 世纪40年代末，它主要用于数值计算，那时的离散元素即为数字．为了进行数值计算而产生了数字计算机．实际上．如果只使用两种状态．数字计算机的性能将更可靠。

因为部件的物理约束和人们的逻辑处理需要．所以数字计算机使用具有两种值的离散数据．这种数据称为二进制。

数字计算机使用只有数字O 和l 的二进制数据系统．一个二进制数字称作位．在数字计算机中用一组二进制表示信息。

应用各种编码技术．一组二进制数字不但可以表示二进制数据，而且还可以表示其他离散符号．如十进制数字或字母表中的字母．例如ASCII是用7 位二进制数字表示一个字符。

通过合理使用二进制组合和各种编码技术．可产生用于完成各种类型计算的一系列指令。

与基数为10 的十进制数相比．二进制数的基数为2 ．使用了数字O 和1 ．与一个二进制数等值的十进制数可以扩展为包含更多位的二进制数．计算机系统是由硬件系统和软件系统组成的，硬件系统由看得见、摸得着的物理设备构成．如磁盘和屏幕。

软件系统是管理计算机的无形“控制”，是运行在计算机中的所有程序，程序是指令的集台，它告诉硬件应该做什么，通常将计算机硬件分为3 个主要部分：输入和输出设备（I/O 设备）、中央处理器（CPU）和存储器。

在第2 章中对硬件部分进行详细介绍．软件可以按其用途分类。

应用软件是为诸如会计、娱乐和工程领域之类的处理任务而设计的，如果玩过计算机游戏或用字处理软件输入入过文件内容．则用户已经具有使用应用软件的体验。

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题（有详细答案）[整理]人教版/bbs/viewthread.php?tid=333&fromuid=9计算机答案合集（组成原理，操作系统...）/bbs/viewthread.php?tid=5204&fromuid=9《应用概率统计》（张国权）版科学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=7334&fromuid=9《C 程序设计》谭浩强编习题答案【khdaw_cola】/bbs/viewthread.php?tid=7836&fromuid=9数据结构（C 语言版）习题答案/bbs/viewthread.php?tid=5200&fromuid=9编译原理 (陈火旺著) 国防工业出版社课后答案/bbs/viewthread.php?tid=2978&fromuid=9微机原理与接口技术第三章 (郭兰英赵祥模著) 清华出版社课后答案_khdaw/bbs/viewthread.php?tid=7347&fromuid=9计算机网络第二版） (冯博琴陈文革著) 高等教育出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=5989&fromuid=9《数据结构》清华大学答案+例程+ppt 课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=9713&fromuid=9软件工程第二版 (张海藩著) 人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=10940&fromuid=9计算机专业英语课后答案【khdaw_cola】/bbs/viewthread.php?tid=7258&fromuid=9数值计算课后答案曾喆昭文卉编著【khdaw_cola】/bbs/viewthread.php?tid=2538&fromuid=9计算机组成与结构(第4 版)习题解答（上）【khdaw_cola】/bbs/viewthread.php?tid=6286&fromuid=9数据结构(c 语言版)习题集答案/bbs/viewthread.php?tid=6096&fromuid=9vb 程序设计（第四版）课后习题答案【khdaw】/bbs/viewthread.php?tid=2994&fromuid=9汇编语言（清华大学出版社）【khdaw】/bbs/viewthread.php?tid=3421&fromuid=9离散数学答案修订版 (耿素云屈婉玲著) 高等教育出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=15917&fromuid=9《编译原理》蒋立源课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=4518&fromuid=9《数字逻辑》（第二版）华中科技大学出版社（欧阳星明）版课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=7332&fromuid=9严蔚敏数据机构实习报告部分答案【khdaw_cola】/bbs/viewthread.php?tid=1036&fromuid=9数值分析部分答案/bbs/viewthread.php?tid=3523&fromuid=9《数字图像处理》（第二版）英文版习题答案（全）【khdaw_lxywyl】/bbs/viewthread.php?tid=5854&fromuid=9C 程序设计(第二版)谭浩强习题解答清华大学出版社课后答案【khdaw_cola】/bbs/viewthread.php?tid=5679&fromuid=9微分几何梅向明第三版黄敬之高等教育出版社课后答案【khdaw】/bbs/viewthread.php?tid=9287&fromuid=9c++语言基础教程 (吕凤葛著) 清华大学出版社课后答案【khdaw】/bbs/viewthread.php?tid=14461&fromuid=9求《数值分析》第五版答案/bbs/viewthread.php?tid=6704&fromuid=9微分几何第三章、第四章 (梅向明著) 高等教育出版社课后答案/bbs/viewthread.php?tid=4721&fromuid=9《计算机组成原理》白中英（第四版•立体化教材）课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=5281&fromuid=9数值分析/bbs/viewthread.php?tid=4245&fromuid=9c 程序设计 (谭浩强著) 清华大学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=7611&fromuid=9《数据库原理》课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=4345&fromuid=9数据结构/bbs/viewthread.php?tid=1624&fromuid=9现代微机原理与接口技术 (杨全胜著) 电子工业出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=11531&fromuid=9《计算机系统组成与体系结构》人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=2852&fromuid=9求数值分析（第４版）李庆扬编/bbs/viewthread.php?tid=6580&fromuid=9java2 实用教程（第三版）【khdaw_cola】/bbs/viewthread.php?tid=7256&fromuid=9数据结构答案【khdaw】/bbs/viewthread.php?tid=2066&fromuid=9人工智能原理及其应用王万森电子工业出版社（2-7 章）【khdaw】/bbs/viewthread.php?tid=9145&fromuid=9JAVA 大学实用教程第二版 (耿祥义张跃平著) 电子工业出版社课后答案【khdaw】/bbs/viewthread.php?tid=15805&fromuid=9Visual C++面向对象编程教程(第2 版) (王育坚著) 清华大学出版社课后答案【khdaw】/bbs/viewthread.php?tid=14404&fromuid=9清华大学出版社计算机网络第4 版中文答案【khdaw_cola】/bbs/viewthread.php?tid=8080&fromuid=9计算机网络第四版 (潘爱民译著) 清华大学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=15080&fromuid=9初等数学研究学习指导 (叶立军著) 华东师范大学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=13939&fromuid=9《复变函数论》张锦豪邱维元版高等教育出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=5919&fromuid=9算法导论原书第二版 (潘金贵顾铁成李成法著) 机械工业出版社课后答案【khdaw】/bbs/viewthread.php?tid=12100&fromuid=9张禾瑞的＜＜近世代数基础＞＞的答案/bbs/viewthread.php?tid=1540&fromuid=9c++程序设计/bbs/viewthread.php?tid=5608&fromuid=9《概率论与统计学》浙大出版社（复习指南）课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=4871&fromuid=9计算机基础课后答案(浙江科学出版社)/bbs/viewthread.php?tid=2014&fromuid=9《C 语言程序设计》张世禄，潘大志，冯天敏电子工业出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=2568&fromuid=9C 语言程序设计(洪维恩)课后答案【khdaw】/bbs/viewthread.php?tid=1955&fromuid=9《计算机组成原理》白中英第四版科学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=6433&fromuid=9微机原理与接口技术第 4 版 (周荷琴，吴秀清著) 中国科学技术大学出版社课后答案【khdaw】/bbs/viewthread.php?tid=15151&fromuid=9数值计算课后答案（清华大学出版）/bbs/viewthread.php?tid=5246&fromuid=9java 程序设计【khdaw_cola】/bbs/viewthread.php?tid=7541&fromuid=9《高等数值分析》清华大学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=5692&fromuid=9数据与计算机通信（第七版） William Stallings 等【khdaw_cola】/bbs/viewthread.php?tid=5201&fromuid=9数值方法第二版 (金一庆陈越著) 机械工业出版社课后答案/bbs/viewthread.php?tid=11539&fromuid=9高等代数北师大高教第三版张和瑞【khdaw】/bbs/viewthread.php?tid=8607&fromuid=9微波技术与天线(第二版) 王新稳李萍李延平编电子工业出版社【khdaw_cola】/bbs/viewthread.php?tid=6534&fromuid=9数据结构（陈慧南编 C＋＋描述）南京邮电大学课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=9506&fromuid=9四川大学出版社编的离散数学教程答案【khdaw_cola】/bbs/viewthread.php?tid=7402&fromuid=9计算机组成原理（白中英版）【khdaw_cola】/bbs/viewthread.php?tid=3243&fromuid=9现代微型计算机与接口教程课后答案杨文显主编寿庆余副主编【khdaw_cola】/bbs/viewthread.php?tid=7964&fromuid=9C 语言程序设计 3-5 章部分程序题答案杨路明北京邮电大学出版社【khdaw】/bbs/viewthread.php?tid=8775&fromuid=9《操作系统》汤子赢西安电子科技大学答案【khdaw_lxywyl】/bbs/viewthread.php?tid=6055&fromuid=9数据库原理与应用教程第二版陈志泊人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=9933&fromuid=9数据结构与算法分析（C++ 第二版）Clifford A. Shaffer 电子工业出版社【khdaw】/bbs/viewthread.php?tid=9211&fromuid=9《程序设计基础》练习题及答案【khdaw_lxywyl】/bbs/viewthread.php?tid=2801&fromuid=9《多媒体技术基础(第2 版)》林福宗清华大学出版社课后参考答案【khdaw】/bbs/viewthread.php?tid=9299&fromuid=9计算机专业英语(含课文、译文、模拟试题、专业英语习题、答案)【khdaw】/bbs/viewthread.php?tid=9364&fromuid=908 版考研概率复习指南答案/bbs/viewthread.php?tid=509&fromuid=9计算机网络（第4 版） (Andrew S.Tanenbaum 著) 清华大学出版社课后答案/bbs/viewthread.php?tid=11361&fromuid=9计算机图形学王汝传 1－4 章人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=10000&fromuid=9计算机网络教程（第3 版）习题答案【khdaw_cola】/bbs/viewthread.php?tid=7777&fromuid=9c++语言程序设计（实验部分）第 3 版(郑莉著) 清华大学出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=15124&fromuid=9数字信号处理学习指导与题解 (丁美玉高西全王军宁著) 电子工业出版社课后答案【khdaw】/bbs/viewthread.php?tid=15168&fromuid=9计算机网络第五版 (谢希仁著) 电子工业出版社课后答案/bbs/viewthread.php?tid=16108&fromuid=9数学物理方程与特殊函数第三版完整 (东南大学数学系王元明著) 高等教育出版社课后答案【khdaw】/bbs/viewthread.php?tid=17311&fromuid=9《操作系统概念》英文版高等教育出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=3491&fromuid=9计算机网络第二版蔡开裕朱培栋徐明（国防科技大学版）【khdaw】/bbs/viewthread.php?tid=9239&fromuid=9《C++语言程序设计教程》吕凤翥人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=8119&fromuid=9电工学第七版下册 (秦曾黄著) 高等教育出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=19778&fromuid=9vfp 表修复工具/bbs/viewthread.php?tid=73&fromuid=9C++语言基础教程吕凤翥人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=10016&fromuid=9数据库及其应用教材课后习题答案_【khdaw_lxywyl】/bbs/viewthread.php?tid=9530&fromuid=9Turbo C 错误信息表/bbs/viewthread.php?tid=70&fromuid=9《微机原理及汇编技术》课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=5684&fromuid=9复变函数答案第四版 (余家荣著) 高等教育出版社课后答案【khdaw】/bbs/viewthread.php?tid=17004&fromuid=9Java 程序设计(第二版) (朱喜福著) 人民邮电出版社课后答案【khdaw_lxywyl】/bbs/viewthread.php?tid=10576&fromuid=9计算机专业英语教程译文（第 4 版） (金志权等主编著) 电子工业出版社课后答案【khdaw】/bbs/viewthread.php?tid=14170&fromuid=9数据结构习题/bbs/viewthread.php?tid=4344&fromuid=9《计算机英语》第２版全书翻译及课后答案_【khdaw_lxywyl】。

《计算机英语(第2版)》参考答案

《计算机英语（第2版）》参考答案注：这里仅给出《计算机英语（第2版）》新增或变化课文的答案，其他未改动课文答案参见《计算机英语（第1版）》原来的答案。

Unit OneSection CPDA Prizefight: Palm vs. Pocket PCI. Fill in the blanks with the information given in the text:1. With DataViz’s Documents To Go, you can view and edit desktop documents on your PDA without converting them first to a PDA-specific ________. (format)2. Both Palm OS and Windows Mobile PDAs can offer e-mail via ________ so that new messages received on your desktop system are transferred to the PDA for on-the-go reading. (synchronization)3. The Windows Mobile keyboard, Block Recognizer, and Letter Recognizer are all ________ input areas, meaning they appear and disappear as needed. (virtual)4. Generally speaking, Windows Mobile performs better in entering information and playing ________ files while Palm OS offers easier operation, more ________ programs, better desktop compatibility, and a stronger e-mail application. (multimedia; third-party)II. Translate the following terms or phrases from English into Chinese and vice versa:1. data field数据字段2. learning curve学习曲线3. third-party solution第三方解决方案4. Windows Media Player Windows媒体播放器5. 开始按钮Start button6. 指定输入区designated input area7. 手写体识别系统handwriting-recognition system8. 字符集character setUnit ThreeSection BLonghorn：The Next Version of WindowsI. Fill in the blanks with the information given in the text:1. NGSCB, the new security architecture Microsoft is developing for Longhorn, splits the OS into two parts: a standard mode and a(n) ________ mode. (secure)2. It is reported that Longhorn will provide different levels of operation that disable the more intensive Aero effects to boost ________ on less capable PCs. (performance)3. With Longhorn’s new graphics and presentation engine, we can create and display Tiles on the desktop, which remind us of the old Active Desktop but are based on ________ instead of ________. (XML; HTML)4. The most talked-about feature in Longhorn so far is its new storage system, WinFS, whichworks like a(n) ________ database. (relational)II. Translate the following terms or phrases from English into Chinese and vice versa:1. search box搜索框2. built-in firewall内置防火墙3. standalone application独立应用程序4. active desktop 活动桌面5. mobile device移动设备6. 专有软件proprietary software7. 快速加载键quick-launch key8. 图形加速器graphics accelerator9. 虚拟文件夹virtual folder10. 三维界面three-dimensional interfaceUnit FourSection CArraysI. Fill in the blanks with the information given in the text:1. Given the array called object with 20 elements, if you see the term object10, you know the array is in ________ form; if you see the term object[10], you know the array is in ________ form. (subscript; index)2. In most programming languages, an array is a static data structure. When you define an array, the size is ________. (fixed)3. A(n) ________ is a pictorial representation of a frequency array. (histogram)4. An array that consists of just rows and columns is probably a(n) ________ array. (two-dimensional)II. Translate the following terms or phrases from English into Chinese and vice versa:1. bar chart条形图2. frequency array频率数组3. graphical representation图形表示4. multidimensional array多维数组5. 用户视图user(’s) view6. 下标形式subscript form7. 一维数组one-dimensional array8. 编程结构programming constructUnit FiveSection BMicrosoft .NET vs. J2EEI. Fill in the blanks with the information given in the text:1. One of the differences between C# and Java is that Java runs on any platform with a Java Virtual ________ while C# only runs in Windows for the foreseeable future. (Machine)2. With .NET, Microsoft is opening up a channel both to ________ in other programming languages and to ________. (developers; components)3. J2EE is a single-language platform; calls from/to objects in other languages are possiblethrough ________, but this kind of support is not a ubiquitous part of the platform. (CORBA)4. One important element of the .NET platform is a common language ________, which runs bytecodes in an Internal Language format. (runtime)II. Translate the following terms or phrases from English into Chinese and vice versa:1. messaging model消息收发模型2. common language runtime通用语言运行时刻（环境）3. hierarchical namespace分等级层次的名称空间4. development community开发社区5. CORBA公用对象请求代理（程序）体系结构6. 基本组件base component7. 元数据标记metadata tag8. 虚拟机virtual machine9. 集成开发环境IDE（integrated development environment）10. 简单对象访问协议SOAP（Simple Object Access Protocol）Unit SixSection ASoftware Life CycleI. Fill in the blanks with the information given in the text:1. The development process in the software life cycle involves four phases: analysis, design, implementation, and ________. (testing)2. In the system development process, the system analyst defines the user, needs, requirements and methods in the ________ phase. (analysis)3. In the system development process, the code is written in the ________ phase. (implementation)4. In the system development process, modularity is a very well-established principle used in the ________ phase. (design)5. The most commonly used tool in the design phase is the ________. (structure chart)6. In the system development process, ________ and pseudocode are tools used by programmers in the implementation phase. (flowcharts)7. Pseudocode is part English and part program ________. (logic)8. While black box testing is done by the system test engineer and the ________, white box testing is done by the ________. (user; programmer)II. Translate the following terms or phrases from English into Chinese and vice versa:1. standard graphical symbol标准图形符号2. logical flow of data标准图形符号3. test case测试用例4. program validation程序验证5. white box testing白盒测试6. student registration system学生注册系统7. customized banking package定制的金融软件包8. software life cycle软件生命周期9. user working environment用户工作环境10. implementation phase实现阶段11. 测试数据test data12. 结构图structure chart13. 系统开发阶段system development phase14. 软件工程software engineering15. 系统分析员system(s) analyst16. 测试工程师test engineer17. 系统生命周期system life cycle18. 设计阶段design phase19. 黑盒测试black box testing20. 会计软件包accounting packageIII. Fill in each of the blanks with one of the words given in the following list, making changes if necessary:development; testing; programmer; chart; engineer; attend; interfacessystem; software; small; userdevelop; changes; quality; board; UncontrolledIV. Translate the following passage from English into Chinese:软件工程是软件开发的一个领域；在这个领域中，计算机科学家和工程师研究有关的方法与工具，以使高效开发正确、可靠和健壮的计算机程序变得容易。

计算机专业英语第二版

计算机专业英语第二版
12.1 Multimedia
Vector graphics use mathematical formulas to recreate the original image. In a vector graphic, the dots are not defined by a row-and-column address; rather they are defined by their spatial relationships to one another. Because their dot components are not restricted to a particular row and column, vector graphics can reproduce images more easily, and they generally look better on most video screens and printers. Common vector graphics formats are Encapsulated Postscript (EPS), Windows Meta (WMF), Hewlett-Packard Graphics Language (HPGL), and Macintosh graphics (PICT). 矢量图形用数学公式来重现原始图像。在矢量图形中，点不是用行和列的地址来定义的；相反，它们是由相互之间的空间关系定义的。因为它们的组成点不严格限制在特定的行和列上，矢量图形可以更容易地复制出图像，而且它们在大多数视屏和打印机上看上去效果更好。常见的矢量图形格式有附录显示格式（EPS）、Windows图元文件格式（WMF）、惠普图形语言(HPGL)和麦金托什图形文件格式。

计算机英语第二版清华第8单元

Computer Science in EnglishDesigned and Programmed ByJiang TongqiangMiao TianshunTeaching Assisted Courseware1Computer Science in EnglishChapter 8An Introduction to Database Systems2Chapter 8I. Pre-reading Questions1. What are the major disadvantages in a file-processing system?(1) Data redundancy and inconsistency.(2) Difficulty in accessing data.(3) Data isolation.(4) Integrity problems.(5) Atomicity problems.(6) Concurrent-access anomalies.(7) Security problems.3Chapter 8I. Pre-reading Questions2. What are the two levels of data independence? Describe each of them.Reference Key:There are two levels of data independence:Physical data independence is the ability to modify the physical schema without causing application programs to be rewritten. Modifications at the physical level are occasionally necessary to improve performance.Logical data independence is the ability to modify the logical schema without causing application programs to be rewritten. Modifications at the logical level are necessary whenever the logical structure of the database is altered (for example, when money-market accounts are added to a banking system).4Chapter 8I. Pre-reading Questions3. What are the functions of the structured query language (SQL)?Reference Key:(1) Data-definition language (DDL).(2) Interactive data-manipulation language (DML).(3) Embedded DML.(4) View definition.(5) Authorization.(6) Integrity.(7) Transaction control.5限制性定语从句that is both convenient and efficient to use in retrieving and storing database information，修饰an environment。

计算机英语实用教程(第二版)课后习题答案

unit1Ex. 1 根据课文内容，回答以下问题。

1. It is primarily responsible for executing instructions.2. Through the monitor.3. All processors, memory modules, plug-in cards, daughterboards, or peripheral devices canbe plugged in those sockets.4. The binary system is used.5. Through clever formatting.6. They are similar to each other. ROM only cannot be altered and does not lose its contentswhen power is removed.7. Mouses.8. Floppy disks c an be removed from their drives and hard disks can’t.9. Because it is faster than many types of parallel port.10. I/O Port.Ex. 2 根据下面的英文解释，写出相应的英文词汇（使用学过的单词、词组或缩略语）。

英文解释词汇A processor whose elements have been miniaturized into one or a few integrated circuits. It is usually used in PC.CPUThe main board of a computer, usually containing the circuitry for the central processing unit, keyboard, and monitor and often having slots for accepting additional circuitry.motherboardA unit of a computer that preserves data for retrieval. memoryA hand-held, button-activated input device that when rolled along a flat surface directs an indicator to move correspondingly about a computer screen, allowing the operator to move the indicator freely, as to select operations or manipulate text or graphics.mouseA set of keys, as on a computer terminal, word processor, typewriter. keyboardA connection point for a peripheral device. portA machine code telling a computer to perform a particular operation. instructionThe information or image displayed at a given time on a monitor, display, or video terminal.screenAn input device, sometimes part of a standard typewriter keyboard,consisting of a separate grid of numerical and function keys arranged for efficient data entry.keypadInformation put into a communications system for transmission or into a computer system for processing.inputEx. 3 把下列句子翻译为中文。

计算机英语(第2版)完整课后答案

练习答案
（The Answers）
PART ONE
Computer Basics
Unit 1
MyComputer
Section A
I．Fill in the blanks with the information given in the text:
1．Charles Babbage; Augusta Ada Byron
Section A
I．Fill in the blanks with the information given in the text:
1．input; output; storage
2．Basic Input Output System
3．flatbed scanners; hand-held scanners
计算机系统包括计算机、外围设备和软件。用于处理数据的电动、电子与机械设备称为硬件。除了计算机本身之外，“硬件”这一术语还指被称为外围设备的组件，这些外围设备可扩大计算机的输入、输出和存储能力。计算机硬件本身在本质上并未提供一件特别有用的智能工具。计算机要想变得有用，就需要有称为软件或计算机程序的一套指令，来告诉计算机如何执行一项特定的任务。计算机在以下情况下变得更加有效：与网络中的其他计算机相连接，以使用户能够共享信息
We can define a computer as a device that accepts input, processes data, stores data, and produces output. According to themodeof processing, computers are either analog ordigital.They can be classified as mainframes, minicomputers, workstations, ormicrocomputers. All else (for example, the age of the machine) being equal, thiscategorizationprovides some indication of the computer’sspeed, size, cost, and abilities.

计算机专业英语课后答案（清华大学出版社第二版）

计算机专业英语课后答案（清华⼤学出版社第⼆版）第⼀章Computer system overview(操作系统概述)I. Pre-reading Questions1.What is a digital computer?The digital computer is a digital system that performs various computational tasks. The word “digital”implies that theinformation in the computer is represented by variables that take a limited number of discrete values. These values are processed internally by components that can maintain a limited number of discrete states. Digital computer is also called electronic computer or computer.2.Are there any differences between the binary number system and the common decimal number system?The decimal digits has ten digits: 0,1,2,,, and 9. The binary number system has two digits: 0 and 1. The binary numbers use a base 2. The decimal digits use a base 10.3.How many types of computers do you know? Name at least four of them. Computers are usually classified into four broad categories: microcomputers, minicomputers, mainframe computers and supercomputers.II Fill in the following blanks.1.Application software is designed to accomplish real-world tasks in fields.2. An 8-bit signed integer can have any value between -128 and 1273. System software controls the computer and enables it to run the hardware and applications software.4. A computer system consists of hardware system and software system .5. List four types of computers: microcomputers, minicomputers , mainframe computers,and s upercomputers .III True or False.1. The software system is the physical equipment that you can see and touch. 1.F2. Typically, a data value is set to zero to represent FALSE and 1 value forTRUE. 2.F3. In the earliest general-purpose computer, most input and output media were magnetic disks. 3. F4.T4. Supercomputers are largest, fastest, and most expensive computer available.5. A computer system consists of hardware system and software system. 5. T IV Matching terms.1、(b) CPUThe processing unit is at the heart of a computer.2、(c) bit A unit of information conveyed by a single binary digit.3、(a) integrated circuit (IC)A complete electronic circuit that packages transistors and other electronic components on a small silicon chip.4、(d) ASCII A system for encoding characters as binary digits.V Translating Terms.(1) coding techniques 编码技术(2) application software 应⽤软件(3) floating point data 浮点数据(4) timesharing 分时，分时技术(5) storage capacities 存储容量VI Choose the best one of the four answers given to fill in each blank.This chapter introduces digital computer, data types, the evolution of computers, and types of computers. 1 is known to all, it’s hard to find a field in 2 computers are not being used. Digital computer, also called electronic computer or computer, is a digital system that 3 various computational tasks. Digital computers use the 4 number system, which has two digits: 0 and 1.By usingvarious coding 5 , groups of bits can be made to represent not only binary numbers 6 other discrete symbols, such as decimal digits or letters of the alphabet. A computer system consists of hardware system and software system. Programs tell the hardware what to do. 7 software is designed to accomplishreal-world tasks in fields such as accounting, entertainment, and engineering. Computers are usually 8 into four broad categories: microcomputers, minicomputers, mainframe computers, and supercomputers. It’s hard to give a 9 definition to each type because computer speeds and storage 10 change rapidly.1. A. As B. It C. As it D. That2. A. what B. which C. where D. when3. A. performs B. carries C. makes D. integrates4. A. decimal B. binary C. Arabian D. American5. A. technique B. Technology C. techniques D. technologies6. A. instead of B. rather than C. but also D. as well7. A. Application B. System C. Word D. Excel8. A. put B. made C. conducted D. classified9. A. precious B. progress C. proceeding D. precise10. A. capacities B. capable C. capabilities D. capacity1.A2. B3. A4. B5. C6. C7. A8. D9. D 1 0.CVII. Translate the following into English.1.By using various coding techniques, groups of bits can be made to represent not only binary numbers but also other discrete symbols.通过应⽤各种编码技术。

计算机专业英语第二版课后翻译答案

Unite 1Section A: 1、artificial intelligence 人工智能2、paper-tape reader 纸带阅读器3、Optical computer 光学计算机4、Neural network 神经网络5、Instruction set 指令集6、Parallel processing 并行处理器7、Difference engine差分机8、Versatile logical element 多用途逻辑元件9、Silicon substrate 硅衬底10、Vaccum tube 真空管11、数据的存储与管理the storage and management of data12、超大规模集成电路large-scale integrated circuit13、中央处理器central processing unit14、个人计算机personal computer15、模拟计算机analog computer16、数字计算机digital computer17、通用计算机general purpose computer18、处理器芯片processor chip19、操作指令operating instructions20、输入设备input devicesSection B1、artificial neural network 人工智能神经网络2、Computer architecture 计算机体系结构3、Robust computer program 健壮的计算机程序4、Human-computer interface 人机接口5、Knowledge representation 知识代表6、数值分析numerical analysis7、程序设计环境programming environment8、数据结构data structure9、存储和检索信息store and retrieve information10、虚拟现实virtual realityUnit 2Section A：1、function key 功能键2、V oice recognition module 声音识别调制器3、Touch-sensitive region 触敏扫描仪4、Address bus 地址总线5、Flatbed scanner 平板扫描仪6、Dot-matrix printer 矩阵式打印机7、Parallel connection 并行连接8、Cathode ray tube 阴极射线管9、Video game 电子游戏10、Audio signal 音频信号11、操作系统operating system12、液晶显示liquid crystal display13、喷墨打印机inkjet printer14、数据总线data bus15、串行连接serial connection16、易失性存储器volatile memory17、激光打印机laser printer18、磁盘存储器floppy disc19、基本输入输出系统basic input/output system20、视频显示器video displaySection B:1、interrupt handler 中断处理程序2、Virtual memory 虚拟内存3、Context switch 上下文转换4、Main memory 主存5、Bit pattern 位模式6、外围设备peripheral device7、进程表process table8、时间片time slice9、图形用户界面graphics user interface10、海量存储器mass storageUnit 3Section A:1、storage register 存储寄存器2、Function statement 函数语句3、Program statement 程序语句4、Object-oriented language 面向对象语言5、Assembly language 汇编语言6、Intermediate language 中间语言7、Relational language 关系语言8、Artificial language 人工语言9、Data declaration 数据声明10、SQL 结构化查询语言11、可执行程序executable program12、程序模块program module13、条件语句conditional statement14、赋值语句assignment statement15、逻辑语言logic statement16、机器语言machine language17、函数式语言functional language18、程序设计语言programming language19、运行计算机程序run a omputer program20、计算机程序员computer programmerSection B1、native code 本机代码2、Header file 头文件3、Multithreaded program 多线程程序4、Java-enabled browser 支持Java的浏览器5、Mallicious code6、机器码machine code7、汇编码assembly code8、特洛伊木马程序trojan9、软件包software package10、类层次class hierarchyUnit 4Section A1、inference engine 推理机2、System call 系统调用3、Compiled language 编译执行的语言4、Parellel computing 并行计算5、Pattern matching 模式匹配6、Memory location 存储单元7、Interpreter program 解释程序8、Library routine 库程序9、Intermediate program 中间程序10、Source file 源文件11、解释执行的语言interpreted language12、设备驱动程序device driver13、源程序source program14、调试程序debugger15、目标代码object code16、应用程序application program17、实用程序utility program18、逻辑程序logic program19、黑盒ink cartridge20、程序的存储与执行storage and execution of program Section B1、Messaging model 通信模式2、Common language runtime 通用语言运行时刻（环境）3、Hierarchical namespace 分层的名称空间4、Development community 开发社区5、CORBA 公共对象请求代理体系结构6、基本组件basic components7、元数据标记metadata token8、虚拟机VM virtual machine9、集成开发环境IDE（intergrated development environment）10、简单对象访问协议SOAP(simple object access protocol) Unit 5Section A1、system specification 系统规范2、Unit testing 单元测试3、Software life cycle 软件的生命周期4、System validation process 系统验证过程5、Evolutionary development process 进化发展过程6、Simple linear model 简单线性模型7、Program unit 程序单元8、Throwaway prototype 一次性使用原型9、Text formatting 文本格式10、System evolution 系统演变11、系统设计范例paradigm for system design12、需求分析与定义Requirements analysis and definition13、探索式编程方法exploratory programming approach14、系统文件编制system documentation15、瀑布模型waterfall model16、系统集成system integration17、商用现成软件commercial off-the-shelf software18、基于组件的软件工程component-based software engineering19、软件维护工具software maintenance tool20、软件复用software reuseSection B1、check box 复选框2、Structured design 结构化设计3、Building block 构建模块4、Database schema 数据库模式5、Radio button 单选按钮6、系统建模技术system modeling techniques7、模型驱动开发MDD(model-driven development)8、数据流程图data flow diagram9、下拉式菜单drop-down10、滚动条scroll barUnit 6Section A1、end user 终端用户2、Atomic operation 原子操作3、Database administrator 数据库管理员4、Relational database model 关系数据库模型5、Local data 本地数据6、Object-oriented database 面向对象的数据库7、Database management system 数据库管理系统8、Entity-relationship model 实体关系模型9、Distributed database 分布式数据库10、Flat file 展开文件11、二维表two-dimensional table12、数据属性data attributes13、数据库对象database object14、存储设备storage device15、数据类型data type16、数据插入与删除insertion and deletion17、层次数据库模型hierarchical18、数据库体系结构database architecture19、关系数据库管理系统ralational database management system20、全局控制总线global control busSection B1、nonvolatile storage system 易失性存储系统2、Equitment malfunction 设备故障3、Wound-wait protocol 损伤等待协议4、Exclusive lock 排它锁5、Database integrity 数据库完整性6、共享锁shared lock7、数据库实现database implementation8、级联回滚cascading rollback9、数据项data item10、分时操作系统time sharing operating system ;Unit 7Section A1、microwave radio 微波无线电2、digital television 数字电视3、DSL 数字用户线路4、analog transmission 模拟传输5、on-screen pointer 屏幕上的指针6、computer terminal 计算机终端7、radio telephone 无线电话8、cellular telephone 蜂窝电话，移动电话，手机9、decentralized network 分散型网络10、wire-based internal network 基于导线的内部网络，有线内部网11、光缆fiber-optic cable12、传真机fax machine13、线通信wireless communications14、点对点通信point-to-point communications15、调制电脉冲modulated electrical impulse16、通信卫星communication(s) satellite17、电报电键telegraph key18、传输媒体transmission medium (或media)19、无绳电话cordless telephone20、金属导体metal conductorSection B1、bit map 位图2、parallel port 并行端口3、direct memory access (DMA) 直接存储器存取4、universal serial bus 通用串行总线5、general-purpose register 通用寄存器6、电路板circuit board7、串行通信serial communication8、数码照相机digital camera9、存储映射输入/输出memory-mapped I/O10、有线电视cable televisionUnit 8Section A1、file server 文件服务器2、carrier sense 载波检测3、Protocol suite 协议族4、Peer-to-peer model 点对点模型5、bus topology network 总线拓扑网络6、inter-machine cooperation 计算机间合作7、Ethernet protocol collection 以太网协作集8、Proprietary network 专有网络9、utility package 实用软件包10、star network 星形网络11、局域网local area network (LAN)12、令牌环token ring13、无线网络wireless network14、封闭式网络closed network15、环形拓扑网络ring topology16、客户/服务机模型client/server model17、网络应用程序network application18、进程间通信interprocess communication19、打印服务机printer server20、广域网wide area networkSection B1、routing path 路由选择通路2、dual-ring topology 双环形拓扑结构3、extended star topology 扩展星形拓扑结构4、backbone network 基干网，骨干网5、mesh topology网络拓扑结构6、同轴电缆coaxial cable7、逻辑拓扑结构logical topology8、无冲突连网环境collision-free networking environment9、树形拓扑结构tree topology10、目的地节点destination nodeUnit 9Section A1、cell phone 蜂窝电话，移动电话，手机2、IP address 网际协议地址，IP地址3、autonomous system 自主系统4、dial-up connection 拨号连接5、network identifier 网络标识符6、binary notation 二进制记数法7、mnemonic name 助记名，缩写名8、Internet-wide directory system 因特网范围的目录系统9、name server 名称服务器10、Internet infrastructure 因特网基础结构11、助记地址mnemonic address12、网吧cyber cafe13、宽带因特网访问broadband Internet access14、顶级域名top-level domain (TLD)15、因特网编址Internet addressing16、点分十进制记数法dotted decimal notation17、因特网服务提供商Internet service provider (ISP)18、专用因特网连接dedicated Internet connection19、主机地址host address20、硬件与软件支持hardware and software support Section B1、incoming message 来报，到来的报文2、application layer 应用层3、utility software 实用软件4、sequence number （顺）序号，序列号5、remote login capabilities 远程登录能力6、端口号port number7、软件例程software routine8、传输层transport layer9、文件传送协议FTP（File Transfer Protocol）10、万维网浏览器Web browserUnit 10Section A1、mailing list 邮件发送清单，邮件列表2、proprietary software 专有软件3、cc line 抄送行4、bcc line 密送行5、forwarded e-mail messages 转发的电子邮件6、e-mail convention 电子邮件常规7、click on an icon 点击图标8、confidential document 密件，秘密文件9、classified information 密级信息10、recovered e-mail message 恢复的电子邮件11、常用情感符commonly used emoticon12、已删除电子邮件deleted e-mail13、电子系统electronic system14、附件行Attachments line15、版权法copyright law16、电子邮件网规e-mail netiquette17、信息高速公路information superhighway18、签名文件signature file19、电子数据表程序spreadsheet program20、文字处理软件word processorSection B1、web-authoring software 网络写作软件2、template generator 模版生成程序3、navigation page 导航页面4、corporate logo 公司标识5、splash page 醒目页面，过渡页6、导航条navigation bar7、节点页面node page8、网站地图site map9、可用性测试usability testing10、图形交换格式gif（Graphics Interchange Format）Unit 11Section A1、customized marketing strategy 定制的营销策略2、B2G transaction 企业对政府交易3、mobile telephone 移动电话4、dot-com bust 网络不景气5、smart card 智能卡，灵巧卡6、digital piracy 数字盗版7、dot-com boom 网络繁荣8、C2C transaction 消费者对消费者交易9、Web auction site 拍卖网站10、fingerprint reader 指纹读取器11、射频识别装置radio-frequency identification (RFID) device12、电子数据交换electronic data interchange (EDI)13、库存管理技术inventory management technology14、知识产权intellectual property15、条形码bar code16、货币兑换currency conversion17、电子图书electronic book18、视网膜扫描仪retina scanner19、个人数字助理personal digital assistant (PDA)20、企业对企业电子商务B2B electronic commerceSection B1、software suite 软件套件2、text box 文本框3、virtual checkout counter 虚拟付款台4、static catalog 静态目录5、browser session 浏览器会话期6、动态目录dynamic catalog7、购物车软件shopping cart software8、供应链supply chain9、企业资源计划软件enterprise resource planning (ERP) software10、税率tax rateUnit 12Section A1、encryption program 加密程序2、deletion command 删除命令3、authorized user 授权的用户4、backup copy 备份5、voltage surge 电压浪涌6、circuit breaker 断路器7、electronic component 电子元件（或部件）8、data-entry error 数据输入错误9、electronic break-in 电子入侵10、power line 电力线，输电线11、检测程序detection program12、电源power source13、破坏性计算机程序destructive computer program14、计算机病毒computer virus15、软件侵权software piracy16、硬盘驱动器hard-disk drive17、病毒检查程序virus checker18、主存储器primary storage19、电子公告板electronic bulletin board20、浪涌电压保护器surge protectorSection B1、phishing attack 网络钓鱼攻击2、graphics card 显（示）卡3、heuristic analysis 试探性分析4、infected file 被感染文件5、virus dictionary 病毒字典6、数据捕获data capture7、恶意软件malicious software8、病毒特征代码virus signature9、防病毒软件antivirus software10、内存驻留程序memory-resident program。

计算机英语第二版清华第6单元

Computer Science in EnglishDesigned and Programmed ByJiang TongqiangMiao TianshunTeaching Assisted Courseware1Computer Science in EnglishChapter 6: Operating Systems2Chapter 6I. Pre-reading Questions1. What is an operating system?Reference Key:The operating system (OS), the most important system software component, consists of the master programs, called the supervisor, that controls the execution of application programs and acts as an interface between the user of a computer and the computer hardware.3Chapter 6I. Pre-reading Questions2. What are functions of operating system?Reference Key:An functions of operating system are consists of the file management, memory management, I/O management, the scheduler, and process management.4Chapter 6I. Pre-reading Questions3. How many operating systems do you know? Name at least four of them.Reference Key:Common operating systems used on microcomputer are DOS, Windows, OS/2, UNIX and Linux.5the most important system software component作the operating system的同位语，起进一步说明的作用。

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专业英语课后习题答案Chapter1I. 1.Application software2. -128 1273. system software4. hardware system software system5.microcomputers, minicomputers, mainframe computers, supercomputersII. 1. false 2. false 3. false 4. true 5.trueIII. (2)CPU(3)bit(1)integrated circuit(IC)(4) ASCIIIV. (1)编码技术(2)应用软件(3) 浮点数据(4)分时(5)存储容量VIII．1.By using various coding techniques, groups of bits can be made to represent not only binary numbers but also other discrete symbols通过应用多种编码技术，一组二进制数字不但可以表示二进制数据，而且还可以表示其它的离散符号2.System software includes not only the complex programs used by technicians to create application software in the first place but also the organizational programs needed to start up the computer and govern its use of other programs.系统软件不仅包括技术人员用于创建应用软件的复杂程序，而且还包括用于启动计算机和提供给其他程序使用的管理程序。

4.Rather than arithmetically or logically manipulating characters, a computer may concatenate strings of characters, replace some characters with others, or otherwise manipulate character strings.计算机能将若干字符连成串，用一些字符代替其他字符或另行处理字符串，而不是用算术方法或逻辑方法处理字符。

5.Software applications like word processing, electronic spreadsheets, database management programs, painting and drawing programs, desktop publishing, and so forth became commercially available, giving more people reasons to use a computer.软件应用，像文字处理、电子表格、数据库管理程序、绘图程序及桌面印刷等进入商业市场，使更多的人去使用计算机。

Chapter2I. 1. the I/O subsystem2. Read Only Memory(ROM)3. SRAM4. I/O interface5. interruptsII. 1. false 2. true 3. false 4. false 5.falseIII. (2)RAM(4)DMA(1)ALU(2)RAMIV. 1.系统总线 2.虚拟存储器 3.计算机体系结构 4. 指令集 5.直接存储器读取VIII.1.By asserting these internal and external control signals in the proper sequence, the control unitcauses the CPU and the rest of the computer to perform the operation needed to correctly process instructions.通过维护内部和外部控制信号的正确顺序，控制单元促使CPU和计算机的其他部件根据正确的处理指令完成操作。

2.In a computer with virtual memory, less-used parts of programs are shifted from RAM to ahard disk and are moved back only when needed.在使用了虚拟存储器的计算机中，很少使用的程序被从RAM移到硬盘中，仅当需要时才被移回来。

3. A technique used to compensate for the mismatch in operating speeds is to employ anextremely fast, small cache between the CPU and main memory whose access time is close to processor logic clock cycle time.用于弥补操作速度的不匹配的一项技术是在CPU和内存之间采用非常快的小容量高速缓冲存储器，它的存取时间接近于处理器的逻辑时钟周期。

4.The data transfer rate of peripherals is usually slower than the transfer rate of the CPU, andconsequently, a synchronization mechanism may be needed.外部设备的数据传输速率通常比CPU的数据传输率更慢，相应的，需要一种同步机制。

5.In some computers the interrupt vector is an address that points to a location in memory wherethe beginning address of the I/O service routine is stored.在一些计算机中，中断向量是一个地址，它指向存储器中存储I/O 服务程序其实地址的单元。

Chatper3I. 1. pipelines 2. the overlapping of computation 3. the length of the vector4. RISC CISC5. instruction stream data streamII. 1. false 2. false 3. true 4. true 5. falseIII. MISD,SIMD,MIMD,SISDIV. 1.并行处理2.流水线处理3.向量处理4.标量处理器5.向下兼容VIII. 1.The purpose of parallel processing is to speed up the computer processing capability and increase its throughout, that is, the amount of processing that can be accomplished during a given interval of time.并行处理的目的是加快计算机的处理速度并增加其吞吐量，所谓吞吐量是在制定规定的时间间隔内所完成的指令。

2.流水线的特征就在于在同一时间里可以在不同的子过程中处理几个不同的运算。

3.为了获得所需要的高性能，有必要采用最快和最可靠的硬件并应用向量处理与并行处理等新技术的革新过程。

4. 总的来说，指令集中的指令书越多，CPU中的传播延迟越大。

5.尽管CISC处理器更为复杂，但这并不一定导致开发成本的增加。

Chapter4I. 1. its syntax, its semantics 2. pseudocode 3. running time 4. maximum 5. listII. a. false b. false c. true d. false e. falseIII. 1.First, divide the original problem into a number of subproblems.Next, solve the subproblems.2. dynamic programmingThe greedy strategy3. a binary tree can be empty, whereas a tree cannot.Each node in a binary tre has exactly two subtrees. Each node in a tree can have any number of subtrees.The subtrees of each node in a binary tree are ordered. The subtree in a tree are unordered. IV. 1.并行算法 2.穷举查找 3.动态规划（动态程序设计）4.双链表 5.二维数组VIII 1.简而言之，算法描述所采用的语言不具有精确的定义或者信息不够详细也会使交流产生问题。

2. 另一种常见的算法结构是当条件为“真”时，循环执行一条语句或者语句序列。

3.在众多算法中，影响运行时间的不仅有不同长度的输入，也包括相同长度的不同输入。

4.因此，动态程序设计是一种从下而上的方法，这种方法通常先解决最小的子问题，并保存其解，再利用这些解对较大问题求解，一步一步地进行直至得到原问题的解。

5.从这种概念性的一维的方式到机器内存中的实际安排的转换是非常直接的。

这些数据可以保存在由程序员想象的相同顺序的24个相邻的内存单元中。

Chapter5I．1. user documentation, operator documentation, programmer documentation2. program3. machine language4. 4GLs5. Natural languagesII. 1. false 2. false 3. false 4. true 5. trueIII1. the first step: problem definitionThe second step: program designThe third step: program codingThe fourth step: program testingThe fifth step: program documentation and maintenance2. machine language、assembly language、high-level language、very-high-level language、natural language3. High-level language programs are compiled, assembly language programs are assembled.High-level languages are platform-independent, the same high-level source code can be compiled to run on different microprocessor and operating systems. Each assembly language is specific to one microprocessor and assembly language programs usually run on only one platform. IV. 1.派生类 2.继承 3.标记语言 4. 超链接5.Java虚拟机VIII. 1.与程序设计相关的每一个人-用户、操作员和程序员都需要文档。

计算机英语(第二版)清华大学出版社 课后习题答案