Chapter 6 Query Processing

合集下载

资讯检索系统测试集之比较

资讯检索系统测试集之比较

Cranfield II (Continued) Cranfield II測試集中相關判斷建立四個步驟
首先請提出查詢問題的建構者對 文件後所附之引用及參考文獻進 行相關判斷
為了避免前述過程仍有遺漏,又 利用文獻耦合的概念計算文件間 之相關性,發掘更多的可能相關 文件。若有兩篇以上的文獻共同 引用了一篇或多篇論文,則稱這 些文獻間具有耦合關係。
< to p > <head> Tipster Topic D escription <num > N um ber: 037 <dom > D om ain: Science and Technology <title> Topic: Identify SA A com ponents <desc> D escription: D ocum ent identifies softw are products w hich adhere to IB M 's S A A standards. <narr> N arrative:
N /A
N /A N /A N /A N /A
N /A
多 主 題 N /A 法 文
N /A
多主題 2 1 日文
N /A
多主題 2 1 日文
Cranfield II
01
比較33種不同索引方式之檢索效益
02
蒐集1400篇有關太空動力學的文件(摘要形式)
03
請每位作者根據這些文件與其當時研究的主題提出問題,經篩選後產 生200餘個查詢問題
4 9 .8 資 訊 科 學 N /A 英 文
3 .1 2 0 .0 1 0 0Biblioteka 7 .22 2 .4

【计算机研究与发展】_query processing_期刊发文热词逐年推荐_20140727

【计算机研究与发展】_query processing_期刊发文热词逐年推荐_20140727

推荐指数 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2010年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
推荐指数 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2011年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
推荐指数 3 1 1 1 1 1 1 1 1 1 1 1 1
2007年 序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
科研热词 网内连接 查询处理层 查询处理 查询代数 最小相关连通子树 数据立方体 数据模型 地理数据服务 图结构 参考点 动态更新 关键字查询 代价模型 xml数据流 xml skyline prmsj icdbs gml dim dcfp
科研热词 xml 高维数据对象 高效 预处理 隐私 连续查询 近似连续范围查询 过滤方法 语义空间 诊断报告 计分 覆盖网络 节点重要性 索引 素数序列标记法 穿行次数 移动计算 离群点检测 模糊 树型xml 查询重写 查询处理 服务质量 最短路径查询 最短距离查询 度量空间 层次聚类 大图 多特征索引 多查询优化 基于位置服务 向量空间 医学病例检索 医学影像 内容与结构查询 关键向量 信息检索 传感器网络 不确定查询处理 不确定数据模型 不确定数据 不确定图 xml压缩 twigstack rq-kd-tree pq-gram p2p系统 knn查询 f&b索引 ccpi

Chapter06_PPT

Chapter06_PPT

10
6.3 Sub Procedures
• Format of a procedure definition
Sub procedure-name(parameter-list) declarations and statements End Sub
• Procedure header
– The first line is known as the procedure header
6.18
2002 Prentice Hall. All rights reserved.
3
6.1 Introduction
• Divide and Conquer
– The best way to develop and maintain a large program is to construct it from small, manageable pieces.
Number 1 2 3 4 5 6 7 8 9 10
Square 1 4 9 16 25 36 49 64 81 100
Outline
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SquareInteger.vb Program Output
2002 Prentice Hall.
All rights reserved.
14
6.4 Function Procedures
• Procedure-name
– Directly follows the Sub keyword – Can be any valid identifier – It is used to call this Sub procedure within the program
• Procedure body

【精】数据库CH(6)(学习资料)

【精】数据库CH(6)(学习资料)

Formal Relational Query LanguagesIn this chapter we study three additional formal relational languages.RelationalAlgebra,tuple relational calculus and domain relational calculus.Of these three formal languages,we suggest placing an emphasis on rela-tional algebra,which is used extensively in the chapters on query processing andoptimization,as well as in several other chapters.The relational calculi generallydo not merit as much emphasis.Our notation for the tuple relational calculus makes it easy to present the concept of a safe query.The concept of safety for the domain relational calcu-lus,though identical to that for the tuple calculus,is much more cumbersomenotationally and requires careful presentation.This consideration may suggestplacing somewhat less emphasis on the domain calculus for classes not focusingon database theory.Exercises6.10Write the following queries in relational algebra,using the universityschema.a.Find the names of all students who have taken at least one Comp.Sci.course.b.Find the ID s and names of all students who have not taken any courseoffering before Spring2009.c.For each department,find the maximum salary of instructors in thatdepartment.You may assume that every department has at least oneinstructor.d.Find the lowest,across all departments,of the per-department maxi-mum salary computed by the preceding query.Answer:4344Chapter6Formal Relational Query Languagesemployee(person name,street,city)works(person name,company name,salary)company(company name,city)manages(person name,manager name)Figure6.22Relational database for Exercises6.2,6.8,6.11,6.13,and6.15a. name(student1takes1 course id(␴dept name=′Comp.Sci.′(course)))Note that if we join student,takes,and course,only students fromthe Comp.Sci.department would be present in the result;studentsfrom other departments would be eliminated even if they had taken aComp.Sci.course since the attribute dept name appears in both studentand course.b. I D,name(student)− I D,name(␴year<2009(student1takes)Note thatSpring is thefirst semester of the year,so we do not need to performa comparison on semester.c.dept name G max(salary)(instructor)d.G min(maxsal)(dept name G max(salary)as maxsal(instructor))6.11Consider the relational database of Figure6.22,where the primary keys areunderlined.Give an expression in the relational algebra to express each ofthe following queries:a.Find the names of all employees who work for“First Bank Corpora-tion”.b.Find the names and cities of residence of all employees who work for“First Bank Corporation”.c.Find the names,street addresses,and cities of residence of all em-ployees who work for“First Bank Corporation”and earn more than$10,000.d.Find the names of all employees in this database who live in the samecity as the company for which they work.e.Assume the companies may be located in several cities.Find all com-panies located in every city in which“Small Bank Corporation”islocated.Answer:(w orks))a. person name(␴company name=“First Bank Corporation”b. person name,city(employee1(w orks)))(␴company name=“First Bank Corporation”Exercises 45c. person name ,street ,city (␴(company name =“First Bank Corporation”∧salar y >10000)w orks 1employee )d. person name (employee 1w orks 1company )e.Note:Small Bank Corporation will be included in each answer.company name (company ÷( city (␴company name =“Small Bank Corporation”(company ))))6.12Using the university example,write relational-algebra queries to find thecourse sections taught by more than one instructor in the following ways:a.Using an aggregate function.b.Without using any aggregate functions.Answer:a.␴instrcnt >1(course id ,section id ,year ,semester G count (∗)as instrcnt (teaches ))b.course id ,section id ,year ,semester (␴I D <>ID2(takes 1␳takes1(ID2,course id ,section id ,year ,semester )(takes )))6.13Consider the relational database of Figure 6.22.Give a relational-algebraexpression for each of the following queries:a.Find the company with the most employees.b.Find the company with the smallest payroll.c.Find those companies whose employees earn a higher salary,on av-erage,than the average salary at First Bank Corporation.Answer:a.t 1←company name G count-distinct (person name )(w orks )t 2←G max (num employees)(␳company strength (company name ,num employees )(t 1))company name (␳t 3(company name ,num employees )(t 1)1␳t 4(num employees )(t 2))b.t 1←company name G sum (salary )(w orks )t 2←G min (payroll )(␳company payroll (company name ,payroll )(t 1)) company name (␳t 3(company name ,payroll )(t 1)1␳t 4(payroll )(t 2))c.t 1←company name G avg (salary )(w orks )t 2←␴company name =“First Bank Corporation”(t 1) t pany name ((␳t 3(company name ,a v g salar y )(t 1))1t 3.a v g salar y >f irst bank .a v g salar y (␳f irst bank (company name ,a v g salar y )(t 2)))6.14Consider the following relational schema for a library:46Chapter6Formal Relational Query Languagesmember(memb no,name,dob)books(isbn,title,authors,publisher)borrowed(memb no,isbn,date)Write the following queries in relational algebra.a.Find the names of members who have borrowed any book publishedby“McGraw-Hill”.b.Find the name of members who have borrowed all books publishedby“McGraw-Hill”.c.Find the name and membership number of members who have bor-rowed more thanfive different books published by“McGraw-Hill”.d.For each publisher,find the name and membership number of mem-bers who have borrowed more thanfive books of that publisher.e.Find the average number of books borrowed per member.Take intoaccount that if an member does not borrow any books,then thatmember does not appear in the borrowed relation at all.Answer:a.t1← isbn(␴publisher=“McGra w−Hill′′(books))name((member1borro w ed)1t1))b.t1← isbn(␴publisher=“McGra w−Hill′′(books))name,isbn(member1borro w ed)÷t1c.t1←member1borro w ed1(␴publisher=“McGra w−Hill′′(books))name(␴countisbn>5((memb no G count-distinct(isbn)as countisbn(t1))))d.t1←member1borro w ed1bookspublisher,name(␴countisbn>5((publisher,memb no G count-distinct(isbn)as countisbn(t1)))6.15Consider the employee database of Figure6.22.Give expressions in tuplerelational calculus and domain relational calculus for each of the followingqueries:a.Find the names of all employees who work for“First Bank Corpora-tion”.b.Find the names and cities of residence of all employees who work for“First Bank Corporation”.c.Find the names,street addresses,and cities of residence of all em-ployees who work for“First Bank Corporation”and earn more than$10,000.Exercises47 d.Find all employees who live in the same city as that in which thecompany for which they work is located.e.Find all employees who live in the same city and on the same streetas their managers.f.Find all employees in the database who do not work for“First BankCorporation”.g.Find all employees who earn more than every employee of“SmallBank Corporation”.h.Assume that the companies may be located in several cities.Find allcompanies located in every city in which“Small Bank Corporation”is located.Answer:a.Find the names of all employees who work for First Bank Corporation:i.{t|∃s∈w orks(t[person name]=s[person name]∧s[company name]=“First Bank Corporation”)} ii.{<p>|∃c,s(<p,c,s>∈w orks∧c=“First Bank Corporation”)} b.Find the names and cities of residence of all employees who work forFirst Bank Corporation:i.{t|∃r∈employee∃s∈w orks(t[person name]=r[person name]∧t[city]=r[city]∧r[person name]=s[person name]∧s[company name]=’First Bank Corporation’)} ii.{<p,c>|∃co,sa,st(<p,co,sa>∈w orks∧<p,st,c>∈employee∧co=“First Bank Corporation”)}c.Find the names,street address,and cities of residence of all employeeswho work for First Bank Corporation and earn more than$10,000perannum:i.{t|t∈employee∧(∃s∈w orks(s[person name]=t[person name]∧s[company name]=“First Bank Corporation”∧s[salar y]> 10000))}ii.{<p,s,c>|<p,s,c>∈employee∧∃co,sa(<p,co,sa>∈w orks∧co=’First Bank Corporation’∧sa>10000)}d.Find the names of all employees in this database who live in the samecity as the company for which they work:48Chapter6Formal Relational Query Languagesi.{t|∃e∈employee∃w∈w orks∃c∈company(t[person name]=e[person name]∧e[person name]=w[person name]∧w[company name]=c[company name]∧e[city]=c[city])}ii.{<p>|∃st,c,co,sa(<p,st,c>∈employee∧<p,co,sa>∈w orks∧<co,c>∈company)}e.Find the names of all employees who live in the same city and on thesame street as do their managers:i.{t|∃l∈employee∃m∈manages∃r∈employee(l[person name]=m[person name]∧m[manager name]=r[person name]∧l[street]=r[street]∧l[city]=r[city]∧t[person name]=l[person name])}ii.{<t>|∃s,c,m(<t,s,c>∈employee∧<t,m>∈manages∧<m,s,c>∈employee)}f.Find the names of all employees in this database who do not workfor First Bank Corporation:If one allows people to appear in the database(e.g.in employee)but notappear in works,the problem is more complicated.We give solutionsfor this more realistic case later.i.{t|∃w∈w orks(w[company name]=“First Bank Corporation”∧t[person name]=w[person name])}ii.{<p>|∃c,s(<p,c,s>∈w orks∧c=“First Bank Corporation”)}If people may not work for any company:i.{t|∃e∈employee(t[person name]=e[person name]∧¬∃w∈w orks(w[company name]=“First Bank Corporation”∧w[person name]=t[person name]))}ii.{<p>|∃s,c(<p,s,c>∈employee)∧¬∃x,y(y=“First Bank Corporation”∧<p,y,x>∈w orks)}g.Find the names of all employees who earn more than every employeeof Small Bank Corporation:i.{t|∃w∈w orks(t[person name]=w[person name]∧∀s∈w orks(s[company name]=“Small Bank Corporation”⇒w[salar y]>s[salar y]))}Exercises49 ii.{<p>|∃c,s(<p,c,s>∈w orks∧∀p2,c2,s2(<p2,c2,s2>∈w orks∨c2=“Small Bank Corporation”∨s>s2))}h.Assume the companies may be located in several cities.Find all com-panies located in every city in which Small Bank Corporation is lo-cated.Note:Small Bank Corporation will be included in each answer.i.{t|∀s∈company(s[company name]=“Small Bank Corporation”⇒∃r∈company(t[company name]=r[company name]∧r[city]=s[city]))}ii.{<co>|∀co2,ci2(<co2,ci2>∈company∨co2=“Small Bank Corporation”∨<co,ci2>∈company)} 6.16Let R=(A,B)and S=(A,C),and let r(R)and s(S)be relations.Write relational-algebra expressions equivalent to the following domain-relational-calculus expressions:a.{<a>|∃b(<a,b>∈r∧b=17)}b.{<a,b,c>|<a,b>∈r∧<a,c>∈s}c.{<a>|∃b(<a,b>∈r)∨∀c(∃d(<d,c>∈s)⇒<a,c>∈s)}d.{<a>|∃c(<a,c>∈s∧∃b1,b2(<a,b1>∈r∧<c,b2>∈r∧b1>b2))}Answer:a. A(␴B=17(r))b.r1sc. A(r)∪(r÷␴B( C(s)))d. r.A((r1s)1c=r2.A∧r.B>r2.B(␳r2(r)))It is interesting to note that(d)is an abstraction of the notoriousquery“Find all employees who earn more than their manager.”LetR=(emp,sal),S=(emp,mgr)to observe this.6.17Repeat Exercise6.16,writing SQL queries instead of relational-algebra ex-pressions.Answer:a.select afrom rwhere b=1750Chapter6Formal Relational Query Languagesb.select a,b,cfrom r,swhere r.a=s.ac.(select afrom r)union(select afrom s)d.select afrom r as r1,r as r2,swhere r1.a=s.a and r2.a=s.c and r1.b>r2.b6.18Let R=(A,B)and S=(A,C),and let r(R)and s(S)be relations.Using the special constant null,write tuple-relational-calculus expressionsequivalent to each of the following:a.r1sb.r1sc.r1sAnswer:a.{t|∃r∈R∃s∈S(r[A]=s[A]∧t[A]=r[A]∧t[B]=r[B]∧t[C]=s[C])∨∃s∈S(¬∃r∈R(r[A]=s[A])∧t[A]=s[A]∧t[C]=s[C]∧t[B]=null)}b.{t|∃r∈R∃s∈S(r[A]=s[A]∧t[A]=r[A]∧t[B]=r[B]∧t[C]=s[C])∨∃r∈R(¬∃s∈S(r[A]=s[A])∧t[A]=r[A]∧t[B]=r[B]∧t[C]=null)∨∃s∈S(¬∃r∈R(r[A]=s[A])∧t[A]=s[A]∧t[C]=s[C]∧t[B]=null)}c.{t|∃r∈R∃s∈S(r[A]=s[A]∧t[A]=r[A]∧t[B]=r[B]∧t[C]=s[C])∨∃r∈R(¬∃s∈S(r[A]=s[A])∧t[A]=r[A]∧t[B]=r[B]∧t[C]=null)}6.19Give a tuple-relational-calculus expression tofind the maximum value inrelation r(A).Answer:{<a>|<a>∈r∧∀<b>∈R a>=b}。

计算机英语词汇

计算机英语词汇

计算机英语词汇1. Hardware (硬件)- CPU (Central Processing Unit, 中央处理器)- RAM (Random Access Memory, 随机存取存储器) - HDD (Hard Disk Drive, 硬盘驱动器)- SSD (Solid State Drive, 固态硬盘)- Monitor (显示器)- Mouse (鼠标)- Keyboard (键盘)- Printer (打印机)- Scanner (扫描仪)2. Software (软件)- OS (Operating System, 操作系统)- Application (应用程序)- Software (软件)- Program (程序)- Driver (驱动程序)- Firmware (固件)3. Internet and Networking (互联网和网络)- Internet (互联网)- Network (网络)- Router (路由器)- Wi-Fi (无线网络)- Ethernet (以太网)- IP (Internet Protocol, 互联网协议)- DNS (Domain Name System, 域名系统)4. Programming (编程)- Code (代码)- Algorithm (算法)- Debugging (调试)- Variable (变量)- Function (函数)- Library (库)- Framework (框架)- Syntax (语法)5. Data and Databases (数据和数据库)- Data (数据)- Database (数据库)- SQL (Structured Query Language, 结构化查询语言) - Data mining (数据挖掘)- Big data (大数据)6. Security (安全)- Virus (病毒)- Malware (恶意软件)- Firewall (防火墙) - Encryption (加密) - Password (密码)。

sql2019 英文版

sql2019 英文版

sql2019 英文版SQL Server 2019是由Microsoft开发的关系型数据库管理系统(RDBMS)。

它是SQL Server系列的最新版本,提供了许多新功能和改进,旨在提高性能、安全性和可扩展性。

SQL Server 2019的英文版提供了全面的英文界面和文档,以满足全球用户的需求。

它支持英文语言环境,并且所有的菜单、选项、对话框和帮助文档都是用英文编写的。

SQL Server 2019英文版的安装过程与其他版本相似。

你需要下载英文版的安装程序,然后按照指示进行安装。

安装完成后,你可以使用英文版的SQL Server Management Studio(SSMS)来管理和操作数据库。

SQL Server 2019英文版提供了许多新功能和改进,包括:1. Big Data Clusters: 可以将SQL Server与Apache Spark和Hadoop集成,实现大数据分析和处理。

2. Intelligent Query Processing: 引入了新的查询优化技术,提高了查询性能和吞吐量。

3. Enhanced Security: 提供了更强大的安全功能,包括数据加密、动态数据掩码和敏感数据分类。

4. Intelligent Performance: 引入了自动调整功能,可以根据工作负载自动调整数据库配置,提高性能和稳定性。

5. Accelerated Database Recovery: 改进了数据库恢复速度,减少了恢复时间。

6. Machine Learning Services: 可以在数据库中运行R和Python脚本,实现机器学习和数据分析。

7. Kubernetes Support: 支持在Kubernetes容器平台上部署和管理SQL Server实例。

总之,SQL Server 2019英文版是一个功能强大的关系型数据库管理系统,它提供了许多新功能和改进,可以满足各种企业和个人用户的需求。

数据库系统概念(英文精编版.第六版)

数据库系统概念(英文精编版.第六版)


Atomicity of updates
Failures
may lead to inconsistencies (1) account_A = account_A – 100 (2) account_B = account_B + 100
Example:

Concurrent access by multiple users
Exercises
Computer users interacts with data in the _______ level A. physical B. logical C. view D. all of the above Application users interact with data in the _______ level. A. physical B. logical C. view D. all of the above How the data are actually stored is called _______ A. Physical level B. Logical level C. View level D. Conceptual level
property is called ( )
A. Data inconsistency C. Data isolation B. Data redundancy D. Data integrity
1.3 View of Data
Hierarchy of Abstraction Levels
Three Abstraction Levels of Data
机械工业出版社
本课程学习内容
关系数据模型 关系数据库语言

引论答案

引论答案

Chapter 1Crssword Puzzle1.people 2、enduser 3、procedures 4、documentfile 5、devicedriver 6、databasefile7、operationsystem 8、hardware 9、Internet 10、modem 11、software 12、information 13、data 14、tablet PCMultiple Choice1-5b b a b d 6-10 d a c d dOpen-ended1.An information system has five parts:people,procedures,software, hardware and data.People:Are end users who use computer to make themselves more productive.Procedures:Specify rules or guidelines for computer operations.Software:Provides step-by-step instructins for computer hardware.Hardware:Processes the data to create information.Data:Consist of unprocessed facts including text,numbers, image and sounds.People are the most important part of any information system.2.System software helps the computer manage its own internal resources.It includes three kindsof programs:operating systems,utilities and device drivers programs.3.Basic applications are the kinds of programs to be considered computer competent.Specialized applications are more narrowly focused on specific disciplines and occupations.Basic applications have some types,such as browser,word processor,spreadsheet ,and so on.Also specialized applications have many types,some of the best known are graphics,audio and video,multimedia,Web authoring ,and artificial intelligence programs.4.There are four types of computers:supercomputers,mainframe computers,minicomputers,andmicrocomputers.Supercomputers are the most powerful type of computer.Mainframe computers occupy specially wired,air-conditioned rooms,they are capable of great processing speeds and data storage.Minicomputers are refrigeratorsized machines.Microcomputers are the least powerful,yet the most widely used and fastest-growing,type of computer.Microcomputers are the most common type of computers.There are four types of microcomputers:Desktop computers,Notebook computers,Handheld computers and PDA.5.Connectivity is the capability of you microcomputer to share the information with othercomputers.Wireless revolution dramatically affect connectivity.A computer network is a communications system connecting two or more computers.Internet is the largest network in the world.The Web provides a multimedia interface to the numerous resources available on the Internet. Chapter2Crossword Puzzle:1、hits2、ISP3、spam4、applets5、surf,search engine6、url7、ftp8、carder 9、attachment 10、B2B 11、Browser 12、filter 13、link 14、downloading ,dsl Multiple Choice:1-5dacca 6-10aaadbMatching:1----I 2----d 3----g 4----e 5---f 6----c 7---h 8---j 9---a 10—bOpen-Ended1.The most common uses of the Internet are the following:communicating,shopping,searching,etertainment,education,and so on.I have participated in almost all of them.I think cmmunicating is the most popular.2. The two most common types of providers are National and wireless.National service providers provide access through standard telephone connections.Wireless service providers provide Internet connections for computers with wireless modems and a wide array of wireless devices.3. A typical e-mail message has three basic elements:header,message,and signature.4. Social networking is the grouping of individuals into specific groups.There are three basic categories of social networking sites:reuniting,friend-of-a-friend,and common interestReuniting sites are designed to connect people who have known one another but have lost touch.Friend-of-a-friend sites are designed to bring together two people who do not know one another but share a common friend.Common interest sites bring together individuals that share common interests or hobbies.5.There are three types of search engines: crawler-based search engines, metasearch engines and specialized search enginesCrawler-based search engines, such as Google, create their listings automatically. Metasearch engines are programs that automatically submit your search request to several engines simultaneously. The metasearch engines receives the results, eliminates duplicates,orders the hits,and then provides the edited list to you .one of the best known is Dogpile.Specialized search engines focus on subject-specific Web sites.such as Environment. Chapter 3Crossword Puzzle1、menu /2、worksheet3、table4、window5、DBMS6、master slide7、dialog box 8、software suite 9、word wrap 10、sort 11、cell 12、range 13、column/ 14、database 15、toolbar 16、document 17、slides 18、queryMultiple Choice1-5dcbdd 6-10abbcbMatching1-5jachb 6-10diegfOpen-Ended1.General-purpose application are widely used in nearly every discipline and occupation.Theyinclude word processors,spreadsheets,database management systems,and presentation graphics.Specialized application includes thousands of other programs that are more narrowly focused on specific disciplines and occupations.2.Most applications use a GUI,use windows to display information,and have menus to presentcommands.3.A formula is an instruction to calculate or process.Functions are prewritten formulas.Aformula related to what-if analysis is called recalculation.4.Presentation graplics are programs that combine a variety of visual objects to createautractive,visually interesting presentation.People in a variety of setting and situations use presentation graphics programs to make their presentations more interesting and professional.5.An integrated package is a single program that provides the functionality of a wordprocessor,spreadsheet,database manager, and more. The primary disadvantage of an integrated package is that the capabilities of each function are not as extensive as in the individual programs.The primary advantages are cost and simplicity.A software suite is a collection of separate application programs bundled together and sold asa group.It is significantly less expensive to buy a suite of applications than to buy eachapplication separatelyChapter 4Crossword Puzzle1、bitmap imag2、vrml3、morphing4、story board5、interactivity6、ACID7、fuzzy logic 8、blog/9、web authoring/10、graphical/ 11、map/ 12、pixels/13、link/ 14、vector image/ 15、ai/ 16、flashMultiple Choice1. a 2、a 3.a 4b 5a 6b 7c 8b 9a 10bMatching1.i 2g 3d 4e 5c 6f 7h 8b 9j10aOpen-Ended1.Graphics are widely used to analyze data and to create professional-looking presentations;Desktop publishers focus on page design and layout and provide greater flexibility;Image editors are programs for creating and editing bitmap images;Illustration programs are used to create and to edit vector images;Image galleries are libraries of electronic images.These images are used for a wide variety of applications from illustrating textbooks to providing visual interest to presentations;Graphics suites are some companies combining their separate graphics programs in groups. 2.Audio editing software allows you to create and edit audio clips,most software allow you toadd audio effects to your tracks. V ideo editing software allows you to reorganize,add effects,and more to your digital video footage.3.Multimedia is the integration of all sorts of media into one presentation.The creation ofinteractive multimedia presentations follows several steps:Plan, Design, Create and Support.4.Creating a site is called Web authoring.A web site design is an interactive multimedia form ofcommunication,designing a web site begins with determining the site’s overall content,the overall site design is commonly represented in a graphical map.More specialized and powerful programs, called Web authoring programs, are typically used to create sophisticated commercial sites.5.The three areas of artificial intelligence are virtual reality ,knowledge-based systems androbotics.V irtual reality is an artificial,or simulated,reality generated in 3-D by acomputer.Knowledge-based systems are a type of artificial intelligence that uses a database to provide assistance to users.Robotics is the field of study concerned with developing and using robots.Chapter 5Crossword Puzzle1.lder .twork server gui fragmented backup,booting warm bootwindows utility suite driver icons tracks nosMultiple Choice1. c 2d 3c 4a 5a 6c 7b 8b 9a 10cMatching1.j 2d 3f 4b 5g 6h 7a 8e 9i 10cOpen-Ended1.System software works with end users,application software,and computer hardware to handlethe majority of technical details.It consists of four types of programs:operating systems,utilities,device drivers and language translators.2.The basic functions of every operating system are :managing computer resources,providing auser interface, and running applications.The three basic operating system categories are :embedded,network and stand-alone.3.Windows is by far the most popular microcomputer operating system today.It comes in avariety of different versions and is designed to run with Intel and Intel-compatible microprocessors.Mac OS is not nearly as widely used as the Windows operating system.It is a powerful,easy-to-use operating system that is popular with professional graphic designs,desktop publishers,and many home users.While Windows,the Mac OS are proprietary operating systems,Linux is not.It is open source software free and available from many sources,including the Web.4.Utilities are specialized programs designed to make computing easier.The five most essentialutilities are :troubleshooting or diagnostic programs,antivirus programs,uninstall programs,backup programs and file compression programs.Unility suites combine several programs into one package.5.A device driver works with the operating system to allow communication between the deviceand the rest of the computer system.Each time the computer system is started, the operating system loads all of the device drivers into memory.Whenever a new device is added to a computer system, a new device driver must be installed before the device can be used.Add Printer Wizard provides step-by-step guidance to select the appropriate printer driver and to install that driver.Windows uses Windows Update to make reinstalling or updating device drivers easily.Chapter 6Crossword Puzzlesystem unit microprocessor bus width memory cable plug and playbyte cache ascii system clockMultiple Choice1. d 2b 3c 4b 5b 6a 7a 8c 9b 10dMatching1.j 2a 3c 4h 5g 6i 7d 8e 9f 10bOpen-Ended1.There are four basic types of system units:Desktop system units,Notebook system units,TabletPC system units ,and Handheld computer system units.Desktop system units typically contain the system’s electronic components and selected secondary storage devices .Notebook system units are portable and much smaller.They contain the electronic components,selected secondary storage devices,and input device .Tablet PC system units are highly protable devices that support the use of a stylus or pen to input commands and data.Handheld computer system units are the smallest and are designed to fit into the palm of one hand.2.The two basic components are:the control unit and the arithmetic-logic unit.Control unit tells the rest of the computer system how to carry out a program’s instructions.Arithmetic-logic unit performs two types of operations---arithmetic and logical.3.There are three types of memory chips:RAM, ROM, and CMOS.RAM chips hold the program and data that the CPU is presently processing,everything in most types of RAM is lost as soon as the microcomputer is turned off or a power failure.ROM chips have programs built into them at the factory. ROM chips are not volatile and cannot be changed by the user.CMOS chips provides flexibility and expandability for a computersystem.It contains essential information that is required every time the computer system is turned on.It is powered by a battery and does not lose its contents when the power is turned.Its contents can be changed to reflect changes in the computer system.4.Five expansion cards are:graphics cards,sound cards,modem cards,NIC and TV tuner cards.Graphics cards connect the system board to the computer’s monitor,they convert the internal electronic sighals to video signals so they can be displayed on the monitor.Sound cards accept audio input from a microphone and convert it into a form that can be processed by the computer,they also convert internal electronic signals to audio signals.Modem cards allow distant computers to communicate with one another by converting electronic signals from within the system unit into electronic signals that can travel over telephone lines and other types of connections.NIC are used to connect a computer to one or more other computers.TV tuner cards can changes the TV signal into one that can be displayed on you monitor.5.Four standard ports are:serial ports,parallel ports, universal serial bus ports and firewire ports.Serial ports are often used to connect a mouse,keyboard,modem,and many other devices to the system unit.Parallel portsa are used to connect external devices that need to send or receive a lot of data over a short distance.USB ports are gradually replacing serial and parallel ports.FireWire ports provide connections to specialized FireWire devices .Chapter 7Crossword Puzzledot pitch keyboard wheel botton mechanical dumb omr refresh rate dpi webcam digital camera toggle printer joystick hdtv upcpixel flexible Multiple Choice1. a 2b 3a 4b 5d 6a 7d 8c 9d 10bMathing1.I 2e 3f 4j5a 6b 7g 8d 9c 10hOpen-Ended1.Input is any data or instructions that are used by a computer.Input devices are hardware usedto translate what people understand into a form that computers can process.2.①There are a widely of different pointing devices including the mouse,joystick,touchscreen,light pen,and stylus.A mouse controls a pointer that is displayed on the monitor.A joystick is the most popular input device for computer games.Y ou control game actions byvarying the pressure,speed,and direction of the joystick.A touch screen is a particular kind of monitor with a clear plastic outer layer.A light pen is a light-sensitive pen-like device.A stylus acts with the computer through handwriting recognition software.②There are three types of scanning devices:optical scanners,card readers,bar code readers,andcharacter and mark recognition devices.An optical scanner accepts documents consisting of text and/or images and converts them to machine readable form.Card readers interpret encoded information.Bar code readers are either handheld wand readers or platform scanners.Character and mark recognition devices are scanners that are able to recognize special characters and marks.③Image capturing devices include digital cameras and digital video cameras.Digital cameras capture still images.Digital video cameras capture motion.④Audio input can take many forms including the human voice and music.V oice recognition systems accept voice commands to control computer operations and to create document.MIDI is a standard for connecting musical instruments to the system unit.3.The most widely used output devices are monitors,printers and audio output.Monitors present visual images of text and graphics.Printers translate information that has been processed by the system unit and present the information on paper.Audio output devices translate audio information from the computer into sounds that people can understand.4. Output is processed data or information.Output devices are hardware used to translateinformation that has been processed by the system unit into a form that humans can understands.5. Devices combines features of input devices such as scanners with features of output deviceslike printers are called combination input and output devices.Combination devices include fax machines,multifunction devices,Internet telephones,and terminals.A fax machine is a standard tool in nearly every office.Multifunction devices combine the capabilities of a scanner,printer,fax,and copying machine.Internet telephone are specialized input and output devices for receiving and sending voice communication.Terminals are input and output device that connect you to a mainframe or other type of computer.Chapter 8Crossword PuzzleTrack Raid access time pits and lands disk caching online storage sector floppy zip cylinder densityMultiple Choice1. a 2a 3c 4a 5c 6a 7c 8d 9d 10cMatching1.g 2a 3j 4i 5d 6b 7h 8c 9f 10eOpen-Ended1.The traditional floppy disk is the 1.44MB 3-inch disk.They have a thin exterior jacker madeof hard plastic to protect the flexible disk inside.The high capacity floppy disks are 3 inches in diameter.They are able to store more information,are thicker,and require special disk drives.2.There are three types of hard disks:internal hard disk,hard-disk cartridge,and hard-disk pack.Three ways to improve the performance of hard disks are disk caching,redundant arrays of inexpensive disks,and file compression/decompression.Disk caching improves hard-disk performance by anticipating data needs.Redundant arrays of inexpensive disks improve performance by expending external storage,improving access speed,and providing reliable storage.File compression and file decompression increase storage capacity by reducing the amount of space required to store data and programs.3.The two most common optical disc formats are CD and DVD.Hi def is the next generation ofoptical discs.①There are four basic types of CDs:read only,write once,rewritable,and Picture and PhotoCDs.Read only CDs cannot be written on or erased by the users,they are used to distribute large databases , references ,and large software application packages.Write once CDs can be written to once,they are used to archive data and to record music download from the Internet.Rewritable CDs are similar to write once CDs except that the disks surface is not permanently altered when data is recorded.Picture and Photo CDs use a special format developed by Eastman Kodak to store digital images.②There are three types of DVDs :read only,write once,and rewriteable.Read only DVDs can provide over two hours of very high-quality video and sound comparable to that found in motion picture theatres.Write once DVDs are typically used to create permanent archives for large amounts of data and to record videos.Rewriteable DVDs are competing rewriteable formats.4.Solid-state storage does not have moving parts.Internet drives use the Internet to store dataand information.Magnetic tape provides sequential access for backup.Solid-state storage is more expensive than the others,it is more reliable and requires less power.Internet drives are low cost and the flexibility to access information from any location using the Internet.Access speed is slower.Magnetic tapes provide slower sequential access.It is an effective and commonly used tool for backing up data.5.Mass storage refers to the storage of large amounts of data in a persisting andmachine-readable fashion. Enterprise storage systems are the computer storage systems designed for large-scale, high-technology environments of the modern enterprises.Mass storage Devices have been described as mass storage include tape libraries, RAID systems, hard disk drives, magnetic tape drives, optical disc drives, magneto-optical disc drives, drum memory, floppy disk drives, punched tape and holographic memoryChapter 9Crossword PuzzleClient bus network network gateway modulation base station packets analog gps topology nic bps Bluetooth Node Bandwidth Dsl hubMultiple Choice1. c 2b 3a 4d 5c 6d 7b 8b 9b 10dMatching1.g 2b 3a 4i 5h 6e 7c 8j 9d 10fOpen-Ended1.①Connectivity is a concept related to using computer networks to link people and resources.②The single most dramatic change in connectivity and communications in the past five yearshas been the widespread use of mobile or wireless telephones.This wireless technology allows individuals to stay connected with one another from almost anywhere at any time,the revolution is just beginning.③Communication systems transmit data from one location to another.Every communicationsystem has four basic elements:sending and receiving devices,communication channel,connection devices,and data transmission specifications.2.①Physical connections use a solid medium to connect sending and receiving devices.These connections include telephone lines,coaxial cable,and fiber-optic cables.Telephone lines consist of twisted pair cable,they have been the standard transmission medium for years for both voice and data.Coaxial cable replaces the multiple wires of telephone lines with a single solid-copper core.Fiber-optic cable transmits data as pulses of light through tiny tubes of glass.②Wireless connections use the air to connect sending and receiving devices.Primary technologies used for wireless connections are infrared,boardcast radio,microwave, and satellite.Infrared uses infrared light waves to communicate over short distances.Broadcast radio communication uses special sending and receiving towers called transceivers.Microwave communication uses high-frequency radio waves.Satellite communication uses sstellites orbiting about 22000 miles above the earth as microwave relay stations.3.The standard protocol for the Internet is TCP/IP.The essential features of this protocolinvolve:identifying sending and receiving devices and reformatting information for transmission across the Internet.Identification:Every computer on the Internet has a unique number address called an IP address,the Internet uses IP addresses to deliver e-mail and to locate Web sites.Reformatting:Information sent or transmitted across the Internet usually travels through numerous interconnected networks.Before the massage is sent,it is reformatted or broken down into small parts called packets,at the receiving end,the packets are reassembled into the correct order.4.The four principal network topologies are star,bus,ring,and hierarchical.In a star network,a number of small computers or peripheral devices are linked to a central unit,all communications pass through this central unit.In a bus network, each device in the network handles its own communication control.In a ring network,each device is connected to two other devices,forming a ring.The hierarchical network consists of several computers linked to a central host computer,other computers are also hosts to other,smaller computers or to peripheral devices.5.The most common network strategies are terminal,client/server,and peer-to-peer systems.①In a terminal network system,processing power is centralized in one large computer.For theend user it is lack of control and flexibility.②Client/server network system use one computer to coordinate and supply services to othernodes on the network.The abiligy of powerful network management software to monitor and control network activities.③In a peer-to-peer network system,nodes have equal anthority and can act as both clients andservers.The networks are inexpensive and easy to install, and they usually work well for smaller systems with fewer than 10 nodes.Chapter 10Crossword PuzzleCookie surge protector mal ware ergononics cracker keystroke logger snoopware hacker history fileMultiple Choice1. a 2a 3d 4b 5a 6c 7d 8c 9b 10dMatching1.g 2h 3f 4a 5b 6c 7j 8d 9i 10eOpen-Endedrge databases raise some serious concerns on privacy:identity theft and mistaken identity.The impact of private networks on privacy have the two parts:The first instance,of firmseavesdropping on employees,has inspired attempts at federal legislation.The second instance,in which online information services screen and reject messages,is a common activity with most commercial services.The internet brings illusion of anonymity,the web creates a history file ,your web activity is monitored is by cookies. To respond to these privacy concerns, a code of fair information practice has been established.2. Computer criminals are of five types:Employees:The largest category of computer criminals consists of those with the easiest access to computers-namely,employees.Outside users:Some suppliers or clients may have access to a company’s computer system.Hackers and crackers:Hackers are people who gain unauthorized access to a computer system for the fun and challenge of it.Crackers do the same thing but for malicious purposes.Organized crime:Members of organized crime groups have discovered that they can use computers just as people in legitimate businesses do,but for illegal purpose.Terrorists:Knowledgeable terrorist groups and hostile governments could potentially crash satellites and wage economic warfare by disrupting navigation and communication systems.3. Some of the principal measures to protect computer security are encryption,restrictingaccess,anticipating disasters,and backing up data.Encryption is the conversion of data into a form, called a ciphertext, that cannot be easily understood by unauthorized people.Corporations have use it for years that some law enforcement agencies are unable to wiretap messages from suspected criminals.Individuals are also using encryption programs to safeguard their private communications.4.Ergonomics is defined as the study of human factors related to things people use.The physical health matters related to computers that have received the most attention recently are the following:eyestrain and headache,back and neck pain,and repetitive strain puter technology creates some irritants that may be counterproductive:noise,and electronic monitoring.5.The basic elements of the Green PC are:System unit ,display, and manufacturing.As a computer user,you can do to help protect the environment are the following:conserve, recycle ,and educate.期末考试题型如下:1. 名词缩写(英文)2. 填空(中文)3. 选择(英文)4. 连线(英文)5. 计算(中文)6. 简答(中文)。

简述信息检索的概念及流程

简述信息检索的概念及流程

简述信息检索的概念及流程## English Answer:### Information Retrieval: Concept and Process.Concept:Information retrieval (IR) is a discipline concerned with the study and design of systems that enable users to locate information relevant to their needs from large collections of documents. It encompasses techniques for indexing and searching documents, as well as evaluating the relevance and effectiveness of search results.Process:The fundamental process of information retrieval typically involves the following stages:1. Document Indexing: Documents are analyzed andrepresented as structured data, known as index terms, which capture their content and features.2. Query Formulation: Users express their information need as a query, which is typically a set of keywords or phrases.3. Query Processing: The query is analyzed and further refined to identify relevant index terms and search strategies.4. Search Execution: Index terms are matched against the document index to retrieve potentially relevant documents.5. Relevance Ranking: Retrieved documents are evaluated and ranked based on their relevance to the user's query.6. Result Presentation: The ranked documents are presented to the user in a user-friendly interface for browsing and selection.### Chinese Answer:信息检索,概念和流程。

Chapter6 ppt

Chapter6 ppt

Background introduction the Fed
Since 1913, the roles and responsibilities of the Federal Reserve System have expanded and its structure has evolved. Its duties today are described as falling into four general areas: 1 Conducting the nation's monetary policy by influencing monetary and credit conditions in the economy in pursuit of maximum employment, stable prices, and moderate long-term interest rates. 2 Supervising and regulating banking institutions to ensure the safety and soundness of the nation's banking and financial system, and protect the credit rights of consumers. 3 Maintaining stability of the financial system and containing systemic risk that may arise in financial markets. 4 Providing financial services to depository institutions, the U.S. government, and foreign official institutions, including playing a major role in operating the nation's payments system.

tpc-c评价指标

tpc-c评价指标

tpc-c评价指标TPC-C是一个用于测试关系型数据库系统的基准测试套件,它包含了一系列的查询操作,涵盖了事务处理、并发控制、索引选择等方面的性能指标。

以下是TPC-C中常用的评价指标:1. QPID:Query Processing Instance Duration,即查询处理实例持续时间。

它是指完成一个事务处理操作所需的总时间,包括事务开始时间、提交时间、回滚时间等。

QPID是TPC-C中最重要的性能指标之一,它可以反映事务处理的效率和吞吐量。

2. TPS:Transactions Per Second,即每秒事务数。

它是指在一秒钟内可以处理的事务数,是衡量系统并发性能的重要指标之一。

3. MAP:Million Active Pages,即百万活跃页面数。

它是指在一秒钟内可以处理的页面数,是衡量数据库系统内存管理能力的重要指标之一。

4. CPU:Central Processing Unit,即中央处理单元。

它是指系统中用于执行事务处理操作的处理器数量,是衡量系统性能的重要指标之一。

5. I/O:Input/Output Operations Per Second,即每秒输入/输出操作数。

它是指系统在一秒钟内完成的输入/输出操作数,是衡量数据库系统磁盘读写能力的重要指标之一。

6. DB Size:Database Size,即数据库大小。

它是指数据库中存储的数据量,是衡量数据库系统数据管理能力的重要指标之一。

7. Locks:Locks,即锁。

它是指数据库系统中用于控制并发访问的锁的数量,是衡量系统并发性能和资源竞争情况的重要指标之一。

这些评价指标都是TPC-C测试中常用的,可以帮助评估数据库系统的性能和可靠性。

在实际应用中,需要根据具体的业务需求和系统特点,选择合适的评价指标进行测试和评估。

Chapter 6 课后答案

Chapter 6 课后答案

Chapter 6 答案Pragmatics1. Define the following terms briefly.(1)pragmatics: a branch of linguistics that studies language in use.(2)deixis: the marking of the orientation or position of entities and situationswith respect to certain points of reference such as the place (here/there) andtime (now/then) of utterance.(3)reference: (in semantics) the relationship between words and the things, actions,events, and qualities they stand for.(4)anaphora: a process where a word or phrase (anaphor) refers back to anotherword or phrase which was used earlier in a text or conversation.(5)presupposition: implicit assumptions about the world required to make anutterance meaningful or appropriate, e.g. “some tea has already been taken”is a presupposition of “Take some more tea!”(6) Speech Act Theory: The theory was proposed by J. L. Austin and has beendeveloped by J. R. Searle. Basically, they believe that language is not only usedto inform or to describe things, it is often used to “do things”, to performacts. In saying “Sorry”, you are performing an act of apology.(7)indirect speech act: an utterance whose literal meaning (location) and intendedmeaning (illocution) are different. For example, Can you pass thesalt? is literally a yes/no question but is usually uttered as a request or politedirective for action.(8)the Cooperative Principle: a principle proposed by the philosopher PaulGrice whereby those involved in communication assume that both partieswill normally seek to cooperate with each other to establish agreed meaning.It is composed of 4 maxims: quality, quantity, relation and manner.(9)the Politeness Principle: politeness is regarded by most interlocutors as ameans or strategy which is used by a speaker to achieve various purposes,such as saving face, establishing and maintaining harmonious social relationsin conversation. This principle requires speakers to “minimize the expressionof impolite beliefs”. It is composed of 6 maxims: Maxims of Tact, Generosity,Approbation, Modesty, Agreement and Sympathy.(10) conversational implicature: the use of conversational maxims in the CooperativePrinciple to produce extra meaning during conversation.2. Deictic expressions: I, now, you, that, here, tomorrow.3. Anaphoric expressions: she, him, it.4. (1) He bought the beer.(2) You have a watch.(3) We bought a car.5.Direct acts: (1)/(5); Indirect acts: (2)/(3)/(4)6.(a) The Maxim of Quality: (1) Do not say what you believe to be false;(2) Do notsay that for which you lack adequate evidence.(b) The Maxim of Quantity: (1) Make your contribution as informative as is required(for the current purpose of the exchange); (2) Do not make your contributionmore informative than is required.(c) The Maxim of Relation: Be relevant.(d) The Maxim of Manner: Be perspicuous (1) Avoid obscurity of expression;(2)Avoid ambiguity; (3) Be brief (avoid unnecessary prolixity); (4) Be orderly.7.The speaker is particularly careful about the maxim of Agreement in PP. The response begins with “well” rather than “no” in order to minimize disagreement between the speaker and hearer.8.It is an indirect speech act. Carol invites Lara to a party, but Lara wants to decline the invitation. To be polite, she doesn’t choose a directrefusal, instead she says “I’ve got an exam tomorrow” as a reasonable excuse to decline the invitation. In this way, she minimizes the expression of impolite beliefs, thus the utterance conforms to PP.。

计算机专业英语词汇表

计算机专业英语词汇表

Content(Key words & terms for chapter 1) 2(Key words & terms for chapter 2) 4(Key words & terms for chapter 3) 6(Key words & terms for chapter 4) 8(Key words & terms for chapter 5) 10(Key words & terms for chapter 6) 12(Key words & terms for chapter 7) 14(Key words & terms for chapter 8) 17(Key words & terms for chapter 1)application software: 应用软件basic application: 基本应用软件communication device: 通讯设备compact disc(CD): 紧凑格式盘computer competency: 计算机技能connectivity: 连通, 连通性data: 数据database file: 数据库文件desktop computer: 台式计算机device driver: 设备驱动程序digital versatile (video) disc: DVD, 数字式通用盘document file: 文档文件end user: 终端用户floppy disk: 软盘handheld computer: 手提电脑hard disk: 硬盘hardware: 硬件information: 信息information system: 信息系统information technology: IT, 信息技术input device: 输入设备Internet: 互联网keyboard: 键盘mainframe computer: 大型机memory: 内存microcomputer: 微型计算机microprocessor: 微处理器midrange computer: 中型机minicomputer: 小型机modem (modulator & demodulator): 调制解调器monitor: 监视器mouse: 鼠标network: 网络notebook computer: 笔记本电脑operating system (OS): 操作系统optical disk: 光盘output device: 输出设备palm computer: 掌上电脑personal digital assistant(PDA): 个人数字化助理presentation file: 演示文稿primary storage: 主存printer: 打印机procedures: 过程/用户文档program: 程序random access memory(RAM): 可读写内存/随机存储器secondary storage: 辅助存储器service program: 服务程序software: 软件special-purpose application: 专用应用软件supercomputer: 超级计算机system software: 系统软件system unit: 系统单元tablet PC: 带手写板的笔记本电脑utility: 工具软件Web: 网wireless revolution: 无线(通讯)革命(Key words & terms for chapter 2)address: 地址Advance Research Project Agency Network (ARPANET): 高级研究项目代理网络applets: Java程序attachment: 附件auction house site: 拍卖网站browser: 浏览器business-to-business (B2B): 商家对商家business-to-consumer (B2C): 商家对消费者cable: 电缆carder: 信用卡盗用者Center for European Nuclear Research (CERN): 欧洲核能源研究中心computer virus: 计算机病毒consumer-to-consumer (C2C): 消费者对消费者dial-up: 拨号digital cash: 电子货币directory search: 目录搜索domain name: 域名downloading: 下载DSL: 美国网络公司e-commerce:电子商务e-learning: 网络学习electronic commerce: 电子商务electronic mail: 电子邮件e-mail: 电子邮件file transfer protocol (FTP): 文件传输协议filter: 过滤器friend: 好友header: 邮件头hit: 点击Hyperlink: 超文本链接Hypertext Markup language (HTML): 超文本标记语言instant messaging (IM): 即时消息Internet: 互联网Internet security suite: 网络安全包Internet Service Provider (ISP) : 网络服务供应商Java: 一种高级语言keyword search: 关键词搜索link: 链接location: 位置(地址)message: 消息metasearch engine: 元搜索引擎national service provider: 国家网络服务中心online: 在线online banking: 网络银行online shopping: 网络购物online stock trading: 网上股票交易person-to-person auction site: 个人对个人拍卖网站plug-in: 插件protocol: 协议search engine: 搜索引擎search service: 搜索服务signature line: 落款social networking: 社会网络spam: 垃圾邮件spam blocker:垃圾邮件拦截器specialized search engine: 专用搜索引擎spider: 代理程序subject: 主题surf: (网上)冲浪top-level domain: 顶层域名Uniform resource locator (URL): 统一资源定位universal instant messenger:通用即时消息软件uploading: 上载Web: 万维网Web auction: 网上拍卖Web-based application:网络应用软件Web-based services:Web 服务Web master:网管Web page: 网页Web utility: 网络工具wireless modem: 无线modemwireless service provider: 无线网络服务中心(Key words & terms for chapter 3)analytical graph: 分析图application software: 应用软件AutoContent wizard: 内容提示向导basic application: 通用软件bulleted list: 带标注段落business suite: 商用套件button:按钮cell: 单元格character effect: 文字效果chart: 图表column:列,栏computer trainer:计算机培训师contextual tab: 文本标签database:数据库database management system (DBMS):数据库管理系统database manager:数据库管理系统design template:设计模板dialog box:对话框document:文本editing:编辑field:域find and replace:查找替换font:字体font size:字号form:表format:格式formula:公式function:函数galleries:图片库grammar checker:语法检查器graphical user interface (GUI):图形用户界面home software:家用版软件home suite:家用套件icons:图标integrated package:集成软件包label:标签master slide:主幻灯片menu:菜单menu bar:菜单栏numbered list:编号序列numeric entry:数字栏personal software:个人版软件personal suite:个人套件pointer:指针presentation graphic:演示图形productivity suite:商用套件query:查询range:一组连续的单元格recalculation:再计算record:记录relational database:关系数据库report:报表ribbons:ribbon 功能区row:行sheet:工作表slide:幻灯片software suite:套装软件sort:排序specialized applications:专用软件specialized suite:专用软件套件speech recognition:语音识别spelling checker:拼写检查器spread sheet:电子表格system software:系统软件table:表格text entry:文本栏thesaurus:辞典toolbar:工具栏user interface:用户界面utility suite:工具套件what-if analysis:推测分析window:窗口word processor:文字处理器word wrap:文字回环/自动换行workbook file:工作表文件worksheet:工作表,电子表格(Key words & terms for chapter 4) animation: 动画artificial intelligence:人工智能artificial reality:虚拟现实audio editing software:声音编辑软件bitmap image:位图图像blog:博客button:按钮clip art:剪贴画desktop publisher:桌面排版软件desktop publishing program:桌面排版软件drawing program:绘画软件expert systems:专家系统flash:动画制作软件fuzzy logic:模糊逻辑graphical map:网页设计结构图graphics suite:图形包HTML editor:HTML编辑软件illustration program: 绘图软件image editors:图像编辑软件image gallery:图像收藏immersive experience:沉浸式体验industrial robot:工业机器人interactivity:交互性knowledge base:知识库knowledge-based system:知识系统link:链接mobile robot:移动机器人Morphing:图像渐变效果multimedia:多媒体multimedia authoring program:多媒体制作软件page layout program: 排版软件perception system robot:感知系统机器人photo editors:照片编辑器pixel:象素raster image:光栅图像robot:机器人robotics:机器人学stock photographs:库存图片story board:网页设计脚本vector :向量/矢量vector illustration:矢量插画vector image:矢量图video editing software:视频编辑软件virtual environment:虚拟环境virtual reality:虚拟现实virtual reality modeling language (VRML):虚拟现实建模语言virtual reality wall:虚拟现实墙VR:虚拟现实Web authoring:网页制作Web authoring programs:网页制作软件Web page editor:网页编辑软件(Key words & terms for chapter 5)Add Printer Wizard: 打印机添加程序antivirus utility :防病毒工具backup:备份backup program:备份软件Boot Camp:BC 软件(使苹果机可使用Windows系统)booting:启动cold boot:冷启动computer support specialist:计算机系统维护专家Dashboard Widgets:苹果机软件包desktop:台式机desktop operating system :桌面操作系统device driver:设备驱动程序diagnostic utility:诊断工具dialog box:对话框Disk cleanup:磁盘清理Disk defragmenter:磁盘碎片整理工具embedded operating system:嵌入式操作系统file:文件file compression program:文件压缩工具folder:文件夹fragmented:碎化graphical user interface (GUI):图形用户界面Help:帮助icon:图标language translator:翻译程序Leopard:苹果机软件LinuxMac OSMac OS Xmenu:菜单multitasking:多任务network operating system (NOS) : 网络操作系统network server:网络服务器One Button Checkup:一种系统工具软件operating system:操作系统platform:平台pointer:指针sectors:扇区software environment:软件环境Spotlight:苹果机软件stand-alone operating system:个人操作系统system software:系统软件tracks:磁道troubleshooting program:故障诊断软件uninstall utility:卸载工具UNIX:一种网络操作系统user interface:用户界面utility:工具软件utility suite:工具软件包virus:病毒warm boot:热启动window:窗口WindowsWindows Add Printer Wizard: Windows打印机添加程序Windows Update: Windows升级程序Windows XP(Key words & terms for chapter 6)AC adapter:交流适配器accelerated graphics port (AGP):加速图形端口analog:模拟信号Arithmetic Logic Unit (ALU):算术逻辑单元arithmetic operation:算术运算ASCII:一种八位键盘符号编码binary coding scheme:二进制编码方案binary system:二进制系统bit:二进制位bus:总线bus line:总线bus width:总线宽度byte:字节cable:电缆cache memory:高速缓冲存储器carrier package:载体Central Processing Unit(CPU):中央处理器chip:芯片clock speed:时钟速度complementary metal-oxide semiconductor (CMOS):互补金属氧化半导体computer technician:计算机技术人员control unit:控制单元coprocessor: 协处理器desktop system unit:台式机系统单元digital:数字式dial-core chip:双核芯片EBCDIC:一种八位键盘符号编码expansion bus:扩展总线expansion card:扩展卡expansion slot:扩展槽FireWire bus:火线总线FireWire port:火线端口flash memory:闪存graphics card:图形卡graphics coprocessor:图形协处理器handheld computer system unit:掌上型计算机系统单元industry standard architecture(ISA):工业标准体系结构Infrared Data Association (IrDA):红外线传输器标准组织integrated circuit:集成电路laptop:膝上型计算机logical operation:逻辑运算memory:内存microprocessor:微处理器modem card: modem卡motherboard:主板Musical Interface Digital Interface (Music Instrument Data Interface, MIDI):乐器数字界面network adapter card:网络适配卡network interface card (NIC):网络接口卡notebook system unit:笔记本电脑系统单元parallel port:并行端口parallel processing:并行处理PC card: PC卡PCI Express : 快速PCIperipheral component interconnect (PCI) :外围设备连接端口Personal digital assistant (PDA):个人数字化助理Plug and Play:即插即用port:端口power supply unit:供电单元processor:处理器random-access memory (RAM):可读写内存read-only memory (ROM):只读内存RFID tag:RFID 标签semiconductor:半导体serial port:串行端口silicon chip:硅芯片slot:插槽smart card:智能卡socket:插口sound card:声卡system board:系统板system bus:系统总线system cabinet:机箱system clock:系统时钟system unit:系统单元tablet PC system unit: tablet PC 系统单元TV tuner card:电视调协卡unicode:十六位字符编码universal serial bus (USB):通用串行总线universal serial bus (USB) port:通用串行总线端口virtual memory:虚拟内存word:字(Key words & terms for chapter 7)active matrix monitor:主动矩阵监视器bar code:条形码bar code reader: 条形码读码器bar code scanner:条形码扫描仪cathode ray tube monitor (CRT):阴极射线管监视器clarity:清晰度combination key:组合键cordless mouse:无线鼠标data projector:数字投影仪digital camera:数字相机digital media player: 数字媒体播放器digital music player: 数字音乐播放器digital video camera:数字摄像机display screen: 显示屏dot-matrix printer:点阵式打印机dot pitch:点距dots-per-inch (dpi):每英寸点数dual-scan monitor:整体扫描显示器dumb terminal:哑终端e-book:电子图书ergonomic keyboard: 保健键盘fax (facsimile) machine:传真机flat-panel monitor:平板显示器flatbed scanner:平板式扫描仪flexible keyboard: 便携式键盘handwriting recognition software: 手写识别软件Headphones:耳机high-definition television (HDTV):高清电视ink-jet printer:喷墨式打印机intelligent terminal:智能终端Internet telephone:网络电话Internet telephony:网络电话传输IP Telephony:网络电话传输joystick:游戏杆keyboard:键盘laser printer:激光打印机light pen:光笔liquid crystal display (LCD):液晶显示器magnetic card reader: 磁卡读卡机Magnetic-ink character recognition (MICR):磁墨符号识别mechanical mouse:机械鼠标monitor:监视器mouse:鼠标mouse pointer:鼠标指针multifunction device (MFD):多功能设备network terminal:网络终端numeric keypad:数字键区optical character recognition (OCR):光学符号识别optical-mark recognition (OMR):光学标记识别optical mouse:光电鼠标optical scanner:光学扫描仪passive-matrix monitor:被动矩阵监视器PDA keyboard: PDA键盘personal laser printer:个人激光打印机photo printer:照片打印机pixel (picture element):象素pixel pitch: 象素点距platform scanner:平板扫描仪plotter:绘图仪pointing stick: 点击杆portable printer: 便携式打印机portable scanner: 便携式扫描仪printer:打印机radio frequency card reader: 无线电频卡读卡机radio frequency identification (RFID):无线电频识别refresh rate:刷新率resolution:分辨率roller ball:滚动球shared laser printer: 共享式激光打印机speakers:扬声器stylus: 尖笔(用于输入)technical writer: 技术文档制作者telephony:电话学,电话传输terminal:终端thermal printer:热感式打印机thin client:网络终端thin film transistor monitor (TFT):主动矩阵显示器toggle key:触发键touch pad:触摸板touch screen:触摸屏trackball:轨迹球traditional keyboard: 传统键盘Universal Product Code (UPC):通用产品代码voice-over IP (VoIP):网络电话传输voice recognition system:语音识别系统wand reader:条形码识别器WebCam:网络摄像机wheel button: (鼠标)滚轮wireless mouse:无线鼠标(Key words & terms for chapter 8)access speed: 读写速度Blu-Ray (BR): 蓝光capacity: 容量CD (compact disc): 紧凑格式盘CD-R (CD-recordable): 可写CD(一次)CD-ROM (compact disc-read only memory): 只读CDCD-ROM jukebox: CD-ROM光盘库CD-RW (compact disc rewritable): 可擦写CD(多次)cylinder: 柱面density: 密度direct access: 直接读写disk caching: 磁盘高速缓冲DVD (digital versatile disc or digital video disc): 数字式通用盘DVD player: DVD播放器DVD-R or DVD+R (DVD recordable): 一次性刻录格式DVD-RAM (DVD-random-access memory):可读写DVDDVD-ROM (digital versatile disc-read only memory): 只读DVD DVD-RW or DVD+RW (DVD rewritable): 可擦写DVD enterprise storage system: 大型存储系统Erasable optical disk: 可擦写光盘file compression: 文件压缩file decompression: 文件解压file server: 文件服务器flash memory card: 闪存卡floppy disk: 软盘floppy disk cartridge: 大容量软盘floppy disk drive (FDD): 软盘驱动器hard disk: 硬盘hard-disk cartridge: 移动式硬盘hard-disk pack: 硬盘盘组HD DVD (high-definition DVD): 高清DVDhead crash: 读写头损坏hi def (high definition): 高清internal hard disk: 内置式硬盘Internet hard drive:在线存储器label: 标签land: 平面magnetic tape: 磁带magnetic tape reel: 磁带卷magnetic tape streamer: 流式磁带mass storage: 海量存储mass storage devices: 海量存储设备media: 介质optical disk: 光盘optical disk drive: 光盘驱动器organizational Internet storage: 高速网络存储PC Card hard disk: PC卡式硬盘pit: 光盘凹陷处primary storage: 主存RAID system: 冗余阵列盘系统redundant arrays of inexpensive disks (RAID): 冗余阵列盘secondary storage: 辅助存储器secondary storage device: 辅助存储设备sector: 扇区sequential access: 顺序读写shutter: 磁盘读写口software engineer: 软件工程师solid-state storage: 固态存储器storage devices: 存储设备tape cartridge: 磁带盒tape library:磁带库track: 磁道USB drive: USB驱动器write-protection notch: 写保护口。

英语考研 胡壮麟版 语言学教程 English Linguistics Chapter 6

英语考研 胡壮麟版 语言学教程 English Linguistics Chapter 6
how frequently the word is used in a given context. 3. Frequency effect: describes the additional ease with which a word is
accessed due to its more frequent usage in the language. 4. Recency effect: describe the additional ease with which a word is
12
2.2 Language comprehension
Mental lexicon: information about the properties of words, retrievable when understanding language
For example, we may use morphological rules
15
The cohort theory hypothesizes that auditory word recognition begins with the formation of a group of words at the perception of the initial sound and proceeds sound by sound with the cohort of words decreasing as more sounds are perceived.
In certain cases, listeners’ knowledge of words can lead to the inhibition of certain phonemes.

dense passage retrieval实践

dense passage retrieval实践

dense passage retrieval实践Dense passage retrieval (DPR) is a method used in information retrieval tasks, specifically in question-answering systems, to efficiently retrieve relevant passages of text from a large corpus. This approach aims to improve the accuracy and speed of retrieving information by using dense vector representations of passages and questions.To implement DPR in practice, several steps need to be taken:1. Data Preprocessing: The first step is to preprocess the data, which involves cleaning and tokenizing the text. This can include removing punctuation, converting to lowercase, and splitting the text into smaller units such as sentences or paragraphs.2. Passage Indexing: In order to retrieve passages efficiently, an index needs to be created. This can be done using techniques like FAISS (Facebook AI Similarity Search) or Annoy (Approximate Nearest Neighbors Oh Yeah). These libraries can efficiently index dense vectors and perform fast similarity searches. Each passage in the corpus is encoded into a dense vector representation using a pre-trained encoder model like BERT or RoBERTa.3. Query Preprocessing: Similar to the data preprocessing step, the query (question) needs to be cleaned and tokenized to prepare it for similarity matching against the passages. The query is also encoded into a dense vector using the same encoder model used for passage encoding.4. Passage Retrieval: Once the query is encoded, the next step is toretrieve the most relevant passages from the index. This is done by computing the cosine similarity between the query vector and all passage vectors in the index. The passages with the highest similarity scores are selected as the relevant passages.5. Post-processing and Ranking: The retrieved passages can be further refined or ranked based on additional criteria. This can include using language models to score the relevance or coherence of the passages, or applying natural language understanding techniques to match the query more accurately to the passages.6. Answer Extraction: Finally, the retrieved passages can be used for answer extraction. This can be done using various methods, such as using named entity recognition models to identify entities in the passages and matching them with the query, or using extractive summarization techniques to extract the most relevant information from the passages.Overall, implementing dense passage retrieval involves data preprocessing, passage indexing, query preprocessing, passage retrieval, post-processing and ranking, and answer extraction. By leveraging dense vector representations and efficient indexing techniques, DPR can significantly improve the accuracy and efficiency of retrieving relevant passages in question-answering systems.。

NVIDIA DOCA DPI MLNX-15-060466 版本1.1 用户指南说明书

NVIDIA DOCA DPI MLNX-15-060466 版本1.1 用户指南说明书

Programming GuideTable of Contents Chapter 1. Introduction (1)1.1. Intended Audience (1)1.2. Changes and New Features in 1.1 (1)Chapter 2. Setup Configuration (2)2.1. Known Issues (2)Chapter 3. DPI Architecture (3)3.1. Signature Database (3)3.2. DPI Queue (3)3.3. Connection Tracking (4)Chapter 4. DPI Initialization and Teardown (5)Chapter 5. Packet Processing (6)5.1. Flow Life Cycle (6)5.2. Enqueueing Packets for Processing (6)5.2.1. Packet Ownership (6)5.2.2. Flow Matching (6)Chapter 6. Performance (8)6.1. Multithreading (8)6.2. RSS and RTE_FLOW (8)Chapter 7. Packet Life Cycle Example (10)Chapter 1.IntroductionDeep packet inspection (DPI) is a method of examining the full content of data packets as they traverse a monitored network checkpoint.DPI provides a more robust mechanism for enforcing network packet filtering as it can be used to identify and block a range of complex threats hiding in network data streams more accurately. This includes:‣Malicious applications‣Malware data exfiltration attempts‣Content policy violations‣Application recognition‣Load balancing1.1. Intended AudienceThis document is intended for software developers writing DPI-based applications such as application recognition (AR), intrusion prevention system (IPS), and intrusion detection system (IDS).The document assumes familiarity with the TCP/UDP stack and data plane development kit (DPDK).1.2. Changes and New Features in 1.1This section provides information regarding the features added and changes made in this software version.‣Added support for IP-based matching‣Added support for non-TCP/UDP streams‣Added support for IP-based signatures‣Added support for action field in signaturesChapter 2.Setup ConfigurationDPI-based application can run either on the host machine, or on the BlueField DPU target.As the DPI leverages the Regular Expressions (RegEx) Engine, users must make sure it is enabled.1.The RegEx engine is enabled by default on the DPU. However, to enable the RegEx on thehost, run the following commands:host$ sudo /etc/init.d/openibd stopdpu$ echo 1 > /sys/class/net/p0/smart_nic/pf/regex_endpu$ cat /sys/kernel/mm/hugepages/hugepages-2048kB/nr_hugepages400 // make sure to allocate 200 additional hugepagesdpu$ echo 600 > /sys/kernel/mm/hugepages/hugepages-2048kB/nr_hugepagesdpu$ systemctl start mlx-regex // To verify the service is properly running, use "systemctl status mlx-regex"host$ sudo /etc/init.d/openibd startNote: Commands with the host$ prompt must be run on the host. Commands with the dpu$ prompt must be run on BlueField.2.Ensure that the BlueField DPU is operating in Ethernet mode, please refer to the DOCAInstallation Guide for more information.3.Ensure that the BlueField DPU is running in embedded CPU function (ECPF) mode(default).Note: Refer to DPU Modes of Operation > "Configuring ECPF Mode from Separated HostMode" under DPU Runtime Guides in the SDK DOCA Developer Zone documentation. 2.1. Known Issues‣The DOCA DPI library only supports inspection of the following protocols:‣http 2.0/1.1/1.0‣TLS/SSL ClientHello and certificate messages‣DNS‣FTP‣TCP/UDP stream-based signatures may detect applications on other protocolsChapter 3.DPI ArchitectureThe following diagram shows how packets are identified by the connection tracking protocol and then injected into the DPI library for processing.3.1. Signature DatabaseThe signature database is compiled into a CDO file by the DPI compiler. The CDO file includes:‣Post-processing table‣Compiled RegEx engine rules‣Other signature informationThe application may load a new database while the DPI is processing packets.For more information on DPI compiler, please refer to the DOCA DPI Compiler document.3.2. DPI QueueA DPI queue is designed to be used by a worker thread. The DPI queue holds the flow’s state. Therefore, all packets from both directions of the flow must be submitted to the same DPI queue "in order". The connection tracking logic will handle out of order packets or retransmission.DPI Architecture 3.3. Connection TrackingFor the DPI library to process cross-packet content, each packet must be injected along with a flow context and a direction. Packets from the same flow direction must be injected "in order". A flow direction is usually represented by a 5-tuple, but it can also be a 3-tuple for other protocols.The connection tracking (CT) logic must handle out of order packets as well as fragmented packets. Once a connection has timed out or terminated, the application must notify the DPI library as well.Chapter 4.DPI Initialization andTeardownBefore enqueueing packets for processing the DPI library must be initialized and loaded with signatures by the main thread:struct doca_dpi_ctx *dpi_ctx = doca_dpi_init(doca_dpi_config);doca_dpi_load_signatures(dpi_ctx, cdo_filename);The following configuration parameters are available:‣nb_queues – number of DPI queues‣max_packets_per_queue – maximum number of packets concurrently processing per queue‣max_sig_match_len – maximum signature length guaranteed to be matched by the DPI libraryFor example: A.*B and max_sig_match_len = 4 guarantees to match AxxB but does not guarantee to match AxxxB.To close the DPI library, the user should call the following function:doca_dpi_destroy(dpi_ctx)Chapter 5.Packet Processing5.1. Flow Life Cycle‣Once a new flow was detected by the connection tracking SW, the user should call doca_dpi_flow_create()‣Every incoming packet classified for this flow should be enqueued by calling doca_dpi_enqueue()‣To poll for the results the application must call doca_dpi_dequeue(). The result will contain matching information (if matched).‣When the connection tracking SW detected that the flow was terminated or aged-out the application should notify the DPI library by calling doca_dpi_flow_destroy()5.2. Enqueueing Packets for ProcessingA call to doca_dpi_enqueue() may reject packets for processing for the following reasons:‣Packet is empty‣DPI queue is full (doca_dpi_dequeue() must be called first)5.2.1. Packet OwnershipFor every mbuf injected, the DPI engine creates an indirect mbuf. This allows the user to free the mbuf at any time after injection. The mbuf mechanism ensures the mbuf returns to the pool only after both the direct and the indirect mbufs are free.If an external attach is used, users must follow the DPDK guidelines forrte_pktmbuf_attach_extbuf() to make sure the mbuf is freed when both the user and the DPI free the mbuf.5.2.2. Flow MatchingA flow may match one or more signatures. The match result will be available to the application on doca_dpi_dequeue(). The DPI library will only report the matched signature with the highest priority. Another way to see the match result for a given flow is to use the function doca_dpi_flow_match_get().Packet ProcessingThe application may query for the application name using doca_dpi_signature_get(). To preserve performance, it is not recommended to call those functions while packets are being processed.It is recommended that the application calls doca_dpi_signatures_get() after loading the database to acquire a copy of the signature names.Chapter 6.Performance6.1. MultithreadingThe DPI library is designed to achieve optimal results in a multi-threaded environment. To achieve best performance, it is recommended that both the packet acquisition and the DPI processing will be done by the same thread.Because some of the DPI work is offloaded to the HW, it is highly recommended that the worker thread will work in a pipeline mode, meaning, it should never wait for a DPI job to be completed but rather go and fetch more packets to be processed. This way the SW can best utilize the CPU while the RegEx accelerator is processing the job.The following pseudocode shows the recommended way to call the DPI library:while(true) {mbufs = rx_burst()foreach mbuf in mbufs {flow_id = connection_tacking(qid, mbuf)if (new flow)doca_dpi_flow_create(qid, flow_id, parsing_info)status = doca_dpi_enqueue(flow_ctx, mbuf, offset)if (status ....)}while(doca_dpi_dequeue(qid, &result) == DOCA_DPI_DEQUEUE_READY)... inspect result ...// At this point processing may not be completed for all packets, so the worker // should continue handling more incoming packets.}6.2. RSS and RTE_FLOWEach flow’s packets must be submitted exclusively to the same queue, for both directions. To achieve that users must either use symmetric RSS or manually (using rte_flow) direct both directions of the flow to the same DPDK queue.PerformanceChapter 7.Packet Life Cycle Example1.The packet is sent to the SFT for processing by calling sft_process_packet() to see ifthe hardware recognizes the flow.2.If the packet is not marked with a zone ID by the HW, the SW must explicitly inform the SFTthe zone of the packet with sft_process_packet_with_zone().3.If the packet is not marked with a flow ID by the HW or the SW, a new flow is created bycalling sft_activate().4.If a new flow ID is assigned by the SFT, doca_dpi_flow_create() must be invoked beforeenqueuing the packet.5.The packet is then processed by the DPI by calling doca_dpi_enqueue().6.If the packet is accepted by the DPI for processing, the result is dequeued by calling doca_dpi_dequeue().7.If a match is found, the result is printed and counted for statistics. The flow then isoffloaded (sent directly to the host) because no further inspection is required.8.To retrieve the match from the DPI engine, doca_dpi_signature_get() allows access tothe sig_data struct which contains the signature ID and string. This action might affect DPI performance.9.When the flow is terminated by the SFT, it should also be destroyed by invokingdoca_dpi_flow_destroy() with the corresponding flow ID.10.Additional statistics can be retrieved using doca_dpi_stat_get().NoticeThis document is provided for information purposes only and shall not be regarded as a warranty of a certain functionality, condition, or quality of a product. NVIDIA Corporation nor any of its direct or indirect subsidiaries and affiliates (collectively: “NVIDIA”) make no representations or warranties, expressed or implied, as to the accuracy or completeness of the information contained in this document and assume no responsibility for any errors contained herein. 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All rights reserved.NVIDIA Corporation | 2788 San Tomas Expressway, Santa Clara, CA 95051。

优化签字流程,提高办事效率的措施

优化签字流程,提高办事效率的措施

优化签字流程,提高办事效率的措施1.确定签字流程的标准步骤,让每个人都清楚该如何操作。

Establish standard procedures for the signing process so that everyone knows exactly what to do.2.简化签字文件的内容,减少不必要的条款和阐述,使得签字流程更加高效。

Simplify the content of the signing documents, reducing unnecessary clauses and explanations to make the signing process more efficient.3.设立签字审核岗位,确保签字文件符合规定,并且避免重复签字。

Establish a signature review position to ensure that the signing documents comply with regulations and to avoid duplicate signatures.4.提供必要的培训和指导,使每个员工都对签字流程了如指掌。

Provide necessary training and guidance so that every employee is well-versed in the signing process.5.利用电子签字工具,减少纸质文件流转的时间和成本。

Utilize electronic signature tools to reduce the time and cost of circulating paper documents.6.将签字文件的审批流程数字化,实现在线审批和签字,提高效率。

Digitize the approval process of signing documents to achieve online approval and signing, improving efficiency.7.精简签字流程的环节,减少不必要的复核和审批人员。

6 Query Operations

6  Query Operations
t
P(ki | R): probability of observing the term k i in the set R of
relevant documents
P(ki | R ): probability of observing the term k i in the set R of
non-relevant documents
10
Relationship
D Cr Dr
Dn
11
12
Formula
• Rocchio,1971 • If Cr is known in advance, then the best query will be represented.
qopt
1 1 C d j N | C | dC d j | Cr | d j r r j r
3
Relevance Feedback Architecture
Query String Revise d Query Query Reformulation
1. Doc1 2. Doc2 3. Doc3 . .
Document corpus
Rankings
IR System
ReRanked Documents
• Bias towards rejecting just the highest ranked of the irrelevant documents:
qm q
d j Dr

d j max nonrelevant (d j )
6-3
: Tunable weight for initial query. : Tunable weight for relevant documents. : Tunable weight for irrelevant document.

算法导论第三版第六章答案英

算法导论第三版第六章答案英

算法导论第三版第六章答案英Chapter6Michelle Bodnar,Andrew LohrDecember31,2016Exercise6.1-1At least2h and at most2h+1?1.Can be seen because a complete binarytree of depth h?1hasΣh?1i=02i=2h?1elements,and the number of elementsin a heap of depth h is between the number for a complete binary tree of depth h?1exclusive and the number in a complete binary tree of depth h inclusive.Exercise6.1-2Write n=2m?1+k where m is as large as possible.Then the heap consists of a complete binary tree of height m?1,along with k additional leaves along the bottom.The height of the root is the length of the longest simple path to one of these kleaves,which must have length m.It is clear from the way we de?ned m that m= lg n .Exercise6.1-3If there largest element in the subtee were somewhere other than the root, it has a parent that is in the subtree.So,it is larger than it’s parent,so,the heap property is violated at the parent of the maximum element in the subtreeExercise6.1-4The smallest element must be a a leaf node.Suppose that node x contains the smallest element and x is not a leaf.Let y denote a child node of x.By the max-heap property,the value of x is greater than or equal to the value of y.Since the elements of the heap are distinct,the inequality is strict.This contradicts the assumption that x contains the smallest element in the heap. Exercise6.1-5Yes,it is.The index of a child is always greater than the index of the parent, so the heap property is satis?ed at each vertex. 1Exercise6.1-6No,the array is not a max-heap.7is contained in position9of the array,so its parent must be in position4,which contains6.This violates the max-heap property.Exercise6.1-7It su?ces to show that the elements with no children are exactly indexed by{ n/2 +1,...,n}.Suppose that we had an i in this range.It’s childeren would be located at2i and2i+1but both of these are≥2 n/2 +2>n and so are not in the array.Now,suppose we had an element with no kids,this means that2i and2i+1are both>n,however,this means that i>n/2.This means that i∈{ n/2 +1,...,n}.Exercise6.2-1271731613101571248902717101613315712489027171016138157124830Exercise6.2-2Algorithm1MIN-HEAPIFY(A,i)1:l=LEF T(i)2:r=RIGHT(i)3:if l≤A.heap?size and A[l]4:smallest=l5:else smallest=i6:end if7:if r≤A.heap?size and A[r]8:smallest=r9:end if10:if smallest=i then11:exchange A[i]with A[smallest]12:MIN-HEAPIFY(A,smallest)13:end ifThe running time of MIN-HEAPIFY is the same as that of MAX-HEAPIFY. Exercise6.2-3The array remains unchanged since the if statement on line line8will be false.2Exercise6.2-4If i>A.heap?size/2then l and r will both exceed A.heap?size so the if statement conditions on lines3and6of the algorithm will never be satis?ed. Therefore largest=i so the recursive call will never be made and nothing will happen.This makes sense because i necessarily corresponds to a leaf node,so MAX–HEAPIFY shouldn’t alter the heap.Exercise6.2-5Iterative Max Heapify(A,i)while il=LEFT(i)r=LEFT(i)largest=iif l≤A.heap-size and A[l]>A[i]thenlargest=lend ifif l≤A.heap-size and A[r]>A[i]thenlargest=rend ifif largest=i thenexchange A[i]and A[largest]elsereturn Aend ifend whilereturn AExercise6.2-6Consider the heap resulting from A where A[1]=1and A[i]=2for 2≤i≤n.Since1is the smallest element of the heap,it must be swapped through each level of the heap until it is a leaf node.Since the heap has height lg n ,MAX-HEAPIFY has worst-case time?(lg n).Exercise6.3-1531710841962295317228419610953192284176109584192231761098451922317610984221953176109842219103176593Exercise6.3-2If we had started at1,we wouldn’t be able to guarantee that the max-heap property is maintained.For example,if the array A is given by[2,1,1,3]then MAX-HEAPIFY won’t exchange2with either of it’s children,both1’s.How-ever,when MAX-HEAPIFY is called on the left child,1,it will swap1with3. This violates the max-heap property because now2is the parent of3.Exercise6.3-3All the nodes of height h partition the set of leaves into sets of size between 2h?1+1and2h,where all but one is size2h.Just by putting all the children of each in their own part of trhe partition.Recall from6.1-2that the heap has height lg(n) ,so,by looking at the one element of this height(the root),we get that there are at most2 lg(n) leaves.Since each of the vertices of height h partitions this into parts of size at least2h?1+1,and all but one corresponds to a part of size2h,we can let k denote the quantity we wish to bound,so,(k?1)2h+k(2h?1+1)≤2 lg(n) ≤n/2sok≤n+2h2h+1+2h+1≤n2h+1≤n2h+1Exercise6.4-14513225717208451320257172845252013717284255201371728425132057172842513208717254413208717252520134871725252013178742525513178742202517135874220252135874172025132587417202513852741720254852713172025845271317202587524131720254752813172025745281317202524578131720255427813172025245781317202542578131720252457813172025Exercise6.4-2We’ll prove the loop invariant of HEAPSORT by induction:Base case:At the start of the?rst iteration of the for loop of lines2-5we have i=A.length.The subarray A[1..n]is a max-heap since BUILD-MAX-HEAP(A)was just called.It contains the n smallest elements,and the empty subarray A[n+1..n]trivially contains the0largest elements of A in sorted order.Suppose that at the start of the i th iteration of of the for loop of lines2-5, the subarray A[1..i]is a max-heap containing the i smallest elements of A[1..n] and the subarray A[i+1..n]contains the n?i largest elements of A[1..n]in sorted order.SinceA[1..i]is a max-heap,A[1]is the largest element in A[1..i]. Thus it is the(n?(i?1))th largest element from the original array since the n?i largest elements are assumed to be at the end of the array.Line3swaps A[1]with A[i],so A[i..n]contain the n?i+1largest elements of the array, and A[1..i?i]contains the i?1smallest elements.Finally,MAX-HEAPIFY is called on A,1.Since A[1..i]was a max-heap prior to the iteration and only the elements in positions1and i were swapped,the left and right subtrees ofnode1,up to node i?1,will be max-heaps.The call to MAX-HEAPIFY will place the element now located at node1into the correct position and restore the5max-heap property so that A[1..i?1]is a max-heap.This concludes the next iteration,and we have veri?ed each part of the loop invariant.By induction, the loop invariant holds for all iterations.After the?nal iteration,the loop invariant says that the subarray A[2..n] contains the n?1largest elements ofA[1..n],sorted.Since A[1]must be the n th largest element,the whole array must be sorted as desired.Exercise6.4-3If it’s already sorted in increasing order,doing the build max heap-max-heap call on line1will takeΘ(n lg(n))time.There will be n iterations of the for loop, each takingΘ(lg(n))time because the element that was at position i was the smallest and so will have lg(n) steps when doing max-heapify on line5.So, it will beΘ(n lg(n))time.If it’s already sorted in decreasing order,then the call on line one will only takeΘ(n)time,since it was already a heap to begin with,but it will still take n lg(n)peel o?the elements from the heap and re-heapify.Exercise6.4-4Consider calling HEAPSORT on an array which is sorted in decreasing order. Every time A[1]is swapped with A[i],MAX-HEAPIFY will be recursively called a number of times equal to the height h of the max-heap containing the elements of positions1through i?1,and has runtime O(h).Since there are2k nodes at height k,the runtime is bounded below bylg ni=12i log(2i)=lg ni=1i2i=2+( lg n ?1)2 lg n =?(n lg n).Exercise6.4-5Since the call on line one could possibly take only linear time(if the input was already a max-heap for example),we will focus on showing that the for loop takes n log n time.This is the case because each time that the last element is placed at the beginning to replace the max element being removed,it has to go through every layer,because it was already very small since it was at the bottom level of the heap before.Exercise6.5-1The following sequence of pictures shows how the max is extracted from the heap.1.Original heap:615 13540126298172.we move the last element to the top of theheap 3.13>9>1so,we swap1and13.4.Since12>5>1,we swap1and12.75.Since6>2>1,we swap1and6.Exercise6.5-2The following sequence of pictures shows how10is inserted into the heap, then swapped with parent nodes until the max-heap property is restored.The node containing the new key is heavily shaded.1.Original heap:815 13540126298172.MAX-HEAP-INSERT(A,10)is called,so we?rst append a node assignedvalue?∞:3.The key value of the new node is updated:4.Since the parent key is smaller than10,the nodes are swapped:95.Since the parent node is smaller than10,the nodes are swapped:Exercise6.5-3Heap-Minimum(A)1:return A[1]Heap-Extract-Min(A)1:if A.heap-size<1then2:Error“heap under?ow”3:end if4:min=A[1]5:A[1]=A[A.heap?size]6:A.heap?size??7:Min-heapify(A,1)8:return minHeap-decrease-key(A,i,key)101:if key?A[i]then2:Error“new key larger than old key”3:end if4:A[i]=key5:while i>1and A[P arent(i)]6:exchange A[i]with A[P arent(i)]7:i=P arent(i)8:end whileMin-Heap-Insert(A,key)1:A.heap?size++2:A[A.heap?size]=∞3:Heap-Decrease-Key(A,A.heap-size,key)Exercise6.5-4If we don’t make an assignment to A[A.heap?size]then it could contain any value.In particular,when we call HEAP-INCREASE-KEY,it might be the case that A[A.heap?size]initially contains a value larger than key,causing an error.By assigning?∞to A[A.heap?size]we guarantee that no error will occur.However,we could have assigned any value less than or equal to key to A[A.heap?size]and the algorithm would still work.Exercise6.5-5Initially,we have a heap and then only change the value at i to make it larger.This can’t invalidate the ordering between i and it’s children,the only other thing that needs to be related to i is that i is less than it’s parent,which may be false.Thus we have the invariant is true at initialization.Then,when we swap i with its parent if it is larger,since it is larger than it’s parent,it must also be larger than it’s sibling,also,since it’s parent was initially above its kids in the heap,we know that it’s parent is larger than it’s kids.The only relation in question is then the new i and it’s parent.At termination,i is the root,so it has no parent,so the heap property must be satis?ed everywhere.Exercise6.5-6Replace A[i]by key in the while condition,and replace line5by“A[i]= A[P ARENT(i)].”After the end of the while loop,add the line A[i]=key. Since the key value doesn’t change,there’s no sense in assigning it until we know where it belongs in the heap.Instead,we only make the assignment of the parent to the child node.At the end of the while loop,i is equal to the position where key belongs since it is either the root,or the parent is at least11key,so we make the assignment.Exercise6.5-7Have a?eld in the structure that is just a count of the total number of elements ever added.When adding an element,use thecurrent value of that counter as the key.Exercise6.5-8The algorithm works as follows:Replace the node to be deleted by the last node of the heap.Update the size of the heap,then call MAX-HEAPIFY to move that node into its proper position and maintain the max-heap property. This has running timeO(lg n)since the number of times MAX-HEAPIFY is recursively called is as most the height of the heap,which is lg n . Algorithm2HEAP-DELETE(A,i)1:A[i]=A[A.heap?size]2:A.heap?size=A.heap?size+13:MAX-HEAPIFY(A,i)Exercise6.5-9Construct a min heap from the heads of each of the k lists.Then,to?nd the next element in the sorted array,extract the minimum element(in O lg(k) time).Then,add to the heap the next element from the shorter list from which the extracted element originally came(also O(lg(k))time).Since?nding the next element in the sorted list takes only at most O(lg(k))time,to? nd the whole list,you need O(n lg(k))total steps.Problem6-1a.They do not.Consider the array A= 3,2,1,4,5 .If we run Build-Max-Heap,we get 5,4,1,3,2 .However,if we run Build-Max-Heap’,we will get 5,4,1,2,3 instead.b.Each insert step takes at most O(lg(n)),since we are doing it n times,weget a bound on the runtime of O(n lg(n)).Problem6-2a.It will su?ce to show how to access parent and child nodes.In a d-ary array,PARENT(i)= i/d ,and CHILD(k,i)=di?d+1+k,where CHILD(k,i) gives the k th child of the node indexed by i.12b.The height of a d-ary heap of n elements is with1of log d n.c.The following is an implementation of HEAP-EXTRACT-MAX for a d-aryheap.An implementation of DMAX-HEAPIFY is also given,which is the analog of MAX-HEAPIFY for d-ary heap.HEAP-EXTRACT-MAX con-sists of constant time operations,followed by a call to DMAX-HEAPIFY.The number of times this recursively calls itself is bounded by the height of the d-ary heap,so the running time is O(d log d n).Note that the CHILD function is meant to be the one described in part(a).Algorithm3HEAP-EXTRACT-MAX(A)for a d-ary heap1:if A.heap?size<1then2:error“heap under?ow”3:end if4:max=A[1]5:A[1]=A[A.heap?size]6:A.heap?size=A.heap?size?17:DMAX-HEAPIFY(A,1)Algorithm4DMAX-HEAPIFY(A,i)1:largest=i2:for k=1to d do3:if CHILD(k,i)≤A.heap?size and A[CHILD(k,i)]>A[i]then4:if A[CHILD(k,i)]>largest then5:largest=A[CHILD(k,i)]6:end if7:end if8:end for9:if largest=i then10:exchange A[i]with A[largest]11:DMAX-HEAPIFY(A,largest)12:end ifd.The runtime of this implementation of INSERT is O(log d n)since the whileloop runs at most as many times as the height of the d-ary array.Note that when we call PARENT,we mean it as de?ned in part(a).e.This is identical to the implementation of HEAP-INCREASE-KEY for2-aryheaps,but with the PARENT function interpreted as in part(a).The run-time is O(log d n)since the while loop runs at most as many times as the height of the d-ary array.13Algorithm5INSERT(A,key)1:A.heap?size=A.heap?size+12:A[A.heap?size]=key3:i=A.heap?size4:while i>1and A[P ARENT(i)5:exchange A[i]with A[P ARENT(i)]6:i=P ARENT(i)7:end whileAlgorithm6INCREASE-KEY(A,i,key)1:if key2:error“new key is smaller than current key”3:end if4:A[i]=key5:while i>1and A[P ARENT(i)6:exchange A[i]with A[P ARENT(i)]7:i=P ARENT(i)8:end whileProblem6-3a.2345 891214 16∞∞∞∞∞∞∞b.For every i,j,Y[1,1]≤Y[i,1]≤Y[i,j].So,if Y[1,1]=∞,we know thatY[i,j]=∞for every i,j.This means that no elements exist.If Y is full,it has no elements labeled∞,in particular,the element Y[m,n]is not labeled ∞.c.Extract-Min(Y,i,j),extracts the minimum value from the young tableau Yobtained by Y [i ,j ]=Y[i +i?1,j +j?1].Note that in running this algorithm,several accesses may be made out of bounds for Y,de? ne these to return∞.No store operations will be made on out of bounds locations.Since the largest value of i+j that this can be called with is n+m,and this quantity must increase by one for each call,we have that the runtime is bounded by n+m.d.Insert(Y,key)Since i+j is decreasing at each step,starts as n+m and isbounded by2below,we know that this program has runtime O(n+m). e.Place the n2elements into a Young Tableau by calling the algorithm frompart d on each.Then,call the algorithm from part c n2to obtain the numbers in increasing order.Both of these operations take time at most2n∈O(n), and are done n2times,so,the total runtime is O(n3)141:min=Y[i,j]2:if Y[i,j+1]=Y[i+1,j]=∞then3:Y[i,j]=∞4:return min5:end if6:if Y[i,j+1]7:Y[i,j]=Y[i,j+1]8:Y[i,j+1]=min9:return Extract-min(y,i,j+1)10:else11:Y[i,j]=Y[i+1,j]12:Y[i+1,j]=min13:return Extract-min(y,i+1,j)14:end if1:i=m,j=n2:Y[i,j]=key3:while Y[i?1,j]>Y[i,j]or Y[i,j?1]>Y[i,j]do 4:if Y[i?1,j]5:Swap Y[i,j]and Y[i,j?1]6:j??7:else8:Swap Y[i,j]and Y[i?1,j]9:i??10:end if11:end while15f.Find(Y,key).Let Check(y,key,i,j)mean to return true if Y[i,j]=key,oth-erwise do nothingi=j=1while Y[i,j]Check(Y,key,i,j)i++end whilewhile i>1and jCheck(i,j)if Y[i,j]j++elsei–end ifend whilereturn false16。

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2014-1-9
集合与包的操作:
关系代数: 集合_没有重复的元组 SQL中: 包——允许有重复的元组
广义笛卡尔积 R×S={t|t=<tr,ts>∧trR∧tsS} r×s元元组的集合 (无重复元组) 与SQL中的FROM子句中的关系 列表对应
水平分割(不消除重复元 素)与SQL中的WHERE子句相
2014-1-9 Chapter 6 10
R
A B C
a 1 10 b 1 20 c 2 10 d 2 35
A =„c‟ I1
S C I2
C D E
10 x 2 20 y 2 30 z 2 40 x 1
<c,2,10> <10,x,2>
check=2? output: <2,x> next tuple: <c,7,15>
2014-1-9
Chapter 6
23
扩展操作符
消除重复
将包转换为集合操作符 与SQL中的DISTINCT相对应
(R) ---DISTINCT
L (R) 表示分组和聚集的结果。
分组和聚集 L
下标 L为: ①关系的一个属性(分组属性); ②关系上的聚集操作符。
分组:GROUP BY 聚集:AVG,SUM,COUNT,
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Chapter 6
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表扫描
• 关系R的元组存放在硬盘上,放在多个数据块中。RDBMS知道哪些块包 含R的元组。 • 将包含关系R元组的数据块依次读入内存的操作称作表扫描 (Table-Scan)。
索引扫描
• 如果R的某个属性上有索引,可以利用该索引得到R的所有元组。可以不 知道哪些数据块包含R。通过索引间接地得到R的所有块,称作索引扫描 (Index-Scan) • 索引扫描的最重要的用途是:进行范围查询或特定值查询。
差: R-S={t|tR∧tS} 交: R∩S={t|tR∧tS}
R - S = {d}
R - S = {a, c, d}
元组t必须在R与S中同时出现。 元组t的出现是在R与S中出现的次数的最小 值。 例: R∩S= = {a, b, c} R∩S= = {a, a, b, c}
Chapter 6 20
选择:
σc(R)={t|tR∧c(t)=‘T’}
水平分割
对应;
与SQL中的SELECT子句相对应
投影:
L(R)
2014-1-9
垂直分割并消除重复元素
①R的属性列
②X→Y:属性X可重命名为Y
③e→z:表达式e重命名为z
Chapter 6 21
包上的选择操作符:
a 0 2 4 2
(R)
b 1 3 5 3
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Chapter 6
3
R A
a
B
1
C
10
S C
10
D
x
E
2
b
c
1
2
20
10
20
30
y
z
2
2
d
e
SQL query :
2
3
35
45
40
50
x
y
1
3
Select B,D From R,S Where R.A = „c‟ AND S.E = 2 AND R.C=S.C
2014-1-9 Chapter 6
2014-1-9
Chapter 6
30
二、扫描表时的排序
需要排序的原因:
(1)SQL语句带ORDER BY子句
(2)关系代数的许多算法要求一个或所有的操作对象是排序的关系, 例如:
sort- merge join
Unique(消除重复) B树的建立 实现排序扫描的方法
2014-1-9
Chapter 6
e 3 45
50 y 3
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Chapter 6
11
Overview of Query Optimization
2014-1-9
Chapter 6
12
SQL query parse parse tree
convert
logical query plan apply laws “improved” l.q.p estimate result sizes l.q.p. +sizes consider physical plans
Chapter 6 Query Processing
2014-1-9
Chapter 6
1
Query Processing
Q Query Plan
Focus: Relational System
2014-1-9
Chapter 6
2
Example
Select B,D From R,S Where R.A = „c‟ And S.E = 2 And R.C=S.C;
statistics
answer execute
Pi
pick best
{(P1,C1),(P2,C2)...}
estimate costs
{P1,P2,…..}
2014-1-9 Chapter 6 13
查询处理器:
第七章
查询 查询编译
查询计划 元数据
第六章
查询执行
数据
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31
• 实现排序扫描的方法,有多种
1. 利用排序属性上的索引扫描(如果有可利用索引);
2. 关系R可以装入内存,先利用表扫描或索引扫描将所有元组读入内
存,然后利用内存排序算法进行排序;
3. 关系R不能装入内存,利用多路归并算法进行排序.
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Chapter 6
32
三、物理操作符计算模型
Chapter 6 7
2014-1-9
Another idea: Plan II
B,D
sR.A = “c”
R
sS.E = 2
S
natural join
2014-1-9
Chapter 6
8
B,D ( sR.A=„c‟(R) ⋈ sS.E=2 (S) )
R
A B C a 1 10 b 1 20 c 2 10
Answer
B 2
D x
4
• How do we execute query?
- Do Cartesian product - Select tuples - Do projection
One idea
2014-1-9
Chapter 6
5
RS
R.A R.B R.C S.C S.D S.E a a . . C . . 1 1 10 10 10 20 x y 2 2
14
查询编译器:
SQL查询 分析查询
查询优化
查询表达式树
逻辑优化
逻辑查询计划树
物理优化
物理查询计划树
执行计划
2014-1-9 Chapter 6
第六章
15
第六章 查询执行
一种查询代数 物理查询计划操作符介绍 数据库操作的一趟算法 嵌套循环连接 基于排序的两趟算法


2014-1-9
MIN,MAX
HAVING:通过在 后附 加选择来表示。
排序: L(R)
下标 L为: 与SQL中的ORDER BY 子句 相对应
排序
2014-1-9
L:关系R中的排序属性列表
Chapter 6
24
例:假设影片关系:
StarsIn( title, year, starname )
检索至少演过三部电影的影星、其中最早的年份: SELECT starName, MIN(year) AS minYear FROM StarIn GROUP BY starName HAVING COUNT(title)>=3; 表示为:
不仅涉及行而且涉及列
算术比较符 辅助专门的关系运算符进行操作 逻辑运算符 辅助专门的关系运算符进行操作
2014-1-9 Chapter 6 19
集合与包的操作:
关系代数: 集合_没有重复的元组 并: R∪S={t|tR∨tS} 消除重复元素 属于R或属于S的元组集合 例:R={a, SQL中: 包_允许有重复元素 不消除重复元素 元组t在R与S中出现次数的和。 例:R={a,
2014-1-9
Chapter 6
17
1.关系代数
一种抽象的查询语言
用对关系的运算来表达查询
2.关系代数运算的三个要素
运算对象:关系 运算结果:关系 运算符:四类
2014-1-9
Chapter 6
18
3. 关系代数运算的分类
集合运算符 将关系看成元组的集合 从关系的“水平”方向即行的角度来进行运算 专门的关系运算符
或仅读满足谓词条件的元组。
扫描表(增加的物理操作符) 扫描表时的排序 物理操作符计算模型 衡量代价cost的参数 扫描操作符的I/O代价 物理操作符的实现 迭代器 iteraor
Chapter 6 28
2014-1-9
一、 Scanning tables(扫描表) 扫描表的常用方法: 表扫描:一次一块的读数据 索引扫描:按索引键读数据
基于散列的两趟算法
基于索引的算法 物理操作符选择与缓冲区管理的关系 使用超过两趟的算法
Chapter 6 16
§6.1一种查询代数 基于集合的关系代数 基于包的关系代数 一、包与集合的区别 在集合中,没有重复的元组 在包中,元组可以重复出现 SQL中的关系实际是包(多重集),而不是集合 二、集合与包的操作
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