密码编码学与网络安全第四版第二章答案翻译
密码编码学与网络安全-原理与实践 课后答案
密码编码学与网络安全原理与实践英文版第四版教学设计
IntroductionCryptography and Network Security: Principles and Practice, 4th Edition is an introductory text on the fundamental principles and practical applications of cryptography and network security. Authored by William Stallings, a renowned expert in the field of computer science, the textbook provides a comprehensive overview of the latest advances in cryptography and network security, including the latest cryptographic algorithms, network security protocols, and security management practices.This document outlines a teaching plan for using Cryptography and Network Security: Principles and Practice, 4th Edition in a classroom setting. The plan includes an overview of the textbook content, suggested teaching objectives and methods, and recommendations for classroom activities and assessments.Textbook OverviewThe textbook is divided into 13 chapters, covering a range of topics related to cryptography and network security. The chapters are organized as follows:1.Overview2.Classical Encryption Techniques3.Block Cipher4.Public-Key Cryptography and RSA5.Key Management and Distribution6.Digital Signatures7.Authentication Protocolswork Security Overview9.Intruders10.Malicious Software11.Firewalls12.Electronic Ml Security13.IP SecurityTeaching ObjectivesThe primary objectives of this class are to:1.Understand the fundamental principles of cryptography andnetwork security.2.Be able to identify and evaluate security threats andvulnerabilities.3.Understand the cryptographic algorithms and protocols usedto secure digital communications.4.Be able to design and implement secure networked systems.5.Develop an understanding of security management practices,including risk assessment and risk mitigation.Teaching MethodsThe primary teaching methods for this course include lectures, group discussions, and hands-on exercises. Lectures will be used to cover the fundamentals of cryptography and network security, including cryptographic algorithms and protocols, as well as network security technologies and practices.Group discussions will facilitate deeper understanding of key concepts and encourage critical thinking about the different approaches to addressing security challenges. Students will be encouraged to share their experiences and perspectives about security issues and to engage in lively debates about topics related to security.Hands-on exercises will provide students with an opportunity to apply the principles and techniques learned in class to real-world scenarios. Students will be required to design and implement secure networked systems, and to evaluate their effectiveness through testing and analysis.Classroom ActivitiesThe following classroom activities are recommended for this course:1.Analysis of security breaches: Students will be assigned toresearch and analyze a recent security breach, and to prepare areport summarizing the breach and suggesting improvements.2.Security audits: Students will work in groups to conduct asecurity audit of a networked system, including vulnerabilityassessments and penetration testing.3.Design and implementation of a secure networked system:Students will work in teams to design and implement a securenetworked system, using the principles and techniques learned in class.4.Guest speakers: Invited guest speakers from industry andgovernment will be invited to share their experiences andperspectives about security issues.AssessmentsThe following assessments are recommended for this course:1.Midterm and final exams: The exams will cover the coursematerial to date, including lectures, discussions, and hands-onexercises.2.Class participation: Students will be evaluated based ontheir class participation, including attendance, contribution to group discussions, and engagement in hands-on exercises.3.Group project: Students will be evaluated based on thequality of their group project, including the design andimplementation of a secure networked system and the evaluation of its effectiveness.ConclusionCryptography and Network Security: Principles and Practice, 4th Edition is an excellent resource for teaching the fundamentals of cryptography and network security. Through lectures, group discussions, and hands-on exercises, students will develop an understanding of the principles and techniques used to secure digital communications and networked systems. By applying these principles and techniques to real-world scenarios, students will be well-equipped to tackle the security challenges of the modern world.。
密码编码学与网络安全课后习题答案全
密码编码学与网络安全课后习题答案全YUKI was compiled on the morning of December 16, 2020密码编码学与网络安全（全）1.1 什么是OSI安全体系结构？OSI安全体系结构是一个架构，它为规定安全的要求和表征满足那些要求的途径提供了系统的方式。
该文件定义了安全攻击、安全机理和安全服务，以及这些范畴之间的关系。
1.2 被动安全威胁和主动安全威胁之间的差别是什么？被动威胁必须与窃听、或监控、传输发生关系。
电子邮件、文件的传送以及用户/服务器的交流都是可进行监控的传输的例子。
主动攻击包括对被传输的数据加以修改，以及试图获得对计算机系统未经授权的访问。
1.4验证：保证通信实体之一，它声称是。
访问控制：防止未经授权使用的资源（即，谁可以拥有对资源的访问，访问在什么条件下可能发生，那些被允许访问的资源做这个服务控制）。
数据保密：保护数据免受未经授权的披露。
数据完整性：保证接收到的数据是完全作为经授权的实体（即包含任何修改，插入，删除或重播）发送。
不可否认性：提供保护反对否认曾参加全部或部分通信通信中所涉及的实体之一。
可用性服务：系统属性或访问和经授权的系统实体的需求，可用的系统资源，根据系统（即系统是可用的，如果它提供服务，根据系统设计，只要用户要求的性能指标它们）。
第二章1.什么是对称密码的本质成分？明文、加密算法、密钥、密文、解密算法。
4.分组密码和流密码的区别是什么？流密码是加密的数字数据流的一个位或一次一个字节。
块密码是明文块被视为一个整体，用来产生一个相同长度的密文块......分组密码每次处理输入的一组分组，相应的输出一组元素。
流密码则是连续地处理输入元素，每次输出一个元素。
6.列出并简要定义基于攻击者所知道信息的密码分析攻击类型。
惟密文攻击：只知道要解密的密文。
这种攻击一般是试遍所有可能的密钥的穷举攻击，如果密钥空间非常大，这种方法就不太实际。
因此攻击者必须依赖于对密文本身的分析，这一般要运用各种统计方法。
密码编码学与网络安全-原理与实践 课后答案
欢迎来到数缘社区。本社区是一个高等数学及密码学的技术性论坛，由山东大学数学院研究 生创办。在这里您可以尽情的遨游数学的海洋。作为站长，我诚挚的邀请您加入，希望大家能一 起支持发展我们的论坛，充实每个版块。把您宝贵的资料与大家一起分享！
Third Edition WILLIAM STALLINGS
Copyright 2002: William Stallings -1-
TABLE OF CONTENTS
Chapter 2: Chaphapter 6: Chapter 7: Chapter 8: Chapter 9: Chapter 10: Chapter 11: Chapter 12: Chapter 13: Chapter 14: Chapter 15: Chapter 16: Chapter 17: Chapter 18: Chapter 19: Chapter 20:
time. A block cipher is one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length. 2.5 Cryptanalysis and brute force. 2.6 Ciphertext only. One possible attack under these circumstances is the brute-force approach of trying all possible keys. If the key space is very large, this becomes impractical. Thus, the opponent must rely on an analysis of the ciphertext itself, generally applying various statistical tests to it. Known plaintext. The analyst may be able to capture one or more plaintext messages as well as their encryptions. With this knowledge, the analyst may be able to deduce the key on the basis of the way in which the known plaintext is transformed. Chosen plaintext. If the analyst is able to choose the messages to encrypt, the analyst may deliberately pick patterns that can be expected to reveal the structure of the key. 2.7 An encryption scheme is unconditionally secure if the ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much ciphertext is available. An encryption scheme is said to be computationally secure if: (1) the cost of breaking the cipher exceeds the value of the encrypted information, and (2) the time required to break the cipher exceeds the useful lifetime of the information. 2.8 The Caesar cipher involves replacing each letter of the alphabet with the letter standing k places further down the alphabet, for k in the range 1 through 25. 2.9 A monoalphabetic substitution cipher maps a plaintext alphabet to a ciphertext alphabet, so that each letter of the plaintext alphabet maps to a single unique letter of the ciphertext alphabet. 2.10 The Playfair algorithm is based on the use of a 5 ¥ 5 matrix of letters constructed using a keyword. Plaintext is encrypted two letters at a time using this matrix. 2.11 A polyalphabetic substitution cipher uses a separate monoalphabetic substitution cipher for each successive letter of plaintext, depending on a key.
(完整版)密码编码学与网络安全第四版第二章答案翻译
第二章2.1什么是对称密码的本质成分？Plaintext, encryption algorithm, secret key, ciphertext, decryption algorithm.明文加密算法密钥密文解密算法2.2 密码算法中两个基本函数式什么？Permutation and substitution.代换和置换P202.3用密码进行通信的两个人需要多少密钥？对称密码只需要一把，非对称密码要两把P202.4 分组密码和流密码的区别是什么？A stream cipher is one that encrypts a digital data stream one bit or one byte at a time. A block cipher is one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length.分组密码每次输入的一组元素，相应地输出一组元素。
流密码则是连续地处理输入元素，每次输出一个元素。
P202.5攻击密码的两种一般方法是什么？Cryptanalysis and brute force.密码分析和暴力破解2.6列出并简要定力基于攻击者所知道信息的密码分析攻击类型。
Ciphertext only. One possible attack under these circumstances is the brute-force approach of trying all possible keys. If the key space is very large, this becomes impractical. Thus, the opponent must rely on an analysis of the ciphertext itself, generally applying various statistical tests to it.Known plaintext.The analyst may be able to capture one or more plaintext messages as well as their encryptions. With this knowledge, the analyst may be able to deduce the key on the basis of the way in which the known plaintext is transformed.Chosen plaintext. If the analyst is able to choose the messages to encrypt, the analyst may deliberately pick patterns that can be expected to reveal the structure of the key.惟密文已知明文选择明文2.7无条件安全密码和计算上安全密码的区别是什么？An encryption scheme is unconditionally secure if the ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much ciphertext is available. An encryption scheme is said to be computationally secure if: (1) the cost of breaking the cipher exceeds the value of the encrypted information, and (2) the time required to break the cipher exceeds the useful lifetime of the information.书本P212.8简要定义Caesar密码The Caesar cipher involves replacing each letter of the alphabet with the letter standing k places further down the alphabet, for k in the range 1 through 25.书本P222.9简要定义单表代换密码A monoalphabetic substitution cipher maps a plaintext alphabet to a ciphertext alphabet, so that each letter of the plaintext alphabet maps to a single unique letter of the ciphertext alphabet.书本P232.10简要定义Playfair密码The Playfair algorithm is based on the use of a 5 5 matrix of letters constructed using a keyword. Plaintext is encrypted two letters at a time using this matrix.书本P262.11单表代换密码和夺标代换密码的区别是什么？A polyalphabetic substitution cipher uses a separate monoalphabetic substitution cipher for each successive letter of plaintext, depending on a key.书本P302.12一次一密的两个问题是什么？1. There is the practical problem of making large quantities of random keys. Any heavily usedsystem might require millions of random characters on a regular basis. Supplying truly random characters in this volume is a significant task.2. Even more daunting is the problem of key distribution and protection. For every message to be sent, a key of equal length is needed by both sender and receiver. Thus, a mammoth key distribution problem exists.书本P332.13什么是置换密码？A transposition cipher involves a permutation of the plaintext letters. 书本P332.14什么是隐写术？Steganography involves concealing the existence of a message.书本P362.7.3习题 2.1a.对b 的取值是否有限制？解释原因。
密码编码学和网络安全
– 因子分解 N=p.q, 从而发觉 ø(N) 和 d – 直接拟定 ø(N) ，然后找出 d – 直接找出 d
• 关键是因子分解
– 改善不大
• Aug-99 曾经使用 GNFS 攻击 130 decimal digits (512) bit
– 算法改善
• 从 “Quadratic Sieve” 到 “Generalized Number Field Sieve”
– 采用1024+ bit RSA 使攻击更困难
• 确保 p, q 具有相同大小且满足其他限制
RSA-定时攻击
• mid-1990’s 提出 • 利用运算过程中旳时间变化
– 密钥中旳0或1，时间不同
• 从化费时间推算出操作数旳大小 • 对策
– 使用固定时间 – 随机延迟 – 运算盲化-取幂操作前将密文与一种随机数相乘
• 线性混合层：确保多轮之上旳 高度扩散。
• 非线性层：具有最优最差-情形 非线性旳S-盒旳并行应用
• 密钥加层：轮密钥简朴地异或 到中间状态上。
AES算法构造
• Ecryption • Decryption
状态、密钥旳表达
• 状态/密钥旳矩阵表达
State
S00 S01 S02 S03 S10 S11 S12 S13 S20 S21 S22 S23 S30 S31 S32 S33
公钥算法应用：鉴别
保密和鉴别
RSA
• RSA措施由三位MIT科学家Rivest、Shamir 和Adleman于1977年提出
• 最著名和使用最广泛旳公钥加密措施 • 基于整数旳有限幂次对素数旳取模 • 使用大整数作为密钥 • 安全性依赖于大数旳因子分解
信息安全基础2(密码编码学与网络安全)
The Phaistos (1700 BC)
Encryption Machines in Early 20th Century
History of Cryptography (cont‘ d)
1949~1976 – Shannon published ―The Communication Theory of Secret Systems‖ in 1949, which indicated cryptography became a formal subject. The development of computer enabled ciphers on complex computing. The security of data is based on the secrecy of secret key instead of cipher algorithm.
– Classical Encryption古典加密 Before computer was invented, cryptography was art more than science. There were some cipher algorithms, encryption machines & simple cryptanalysis密 码分析 ways. The main encryption objects are alphabet character. The security of data is based on the secrecy of algorithms.
Symmetric Cipher Ke=Kd Asymmetric Cipher Ke≠Kd Ke Kd So, make Ke public, keep Kd secret
密码编码学与网络安全课后习题答案全修订稿
密码编码学与网络安全课后习题答案全Document number【SA80SAB-SAA9SYT-SAATC-SA6UT-SA18】密码编码学与网络安全（全）什么是OSI安全体系结构？OSI安全体系结构是一个架构，它为规定安全的要求和表征满足那些要求的途径提供了系统的方式。
该文件定义了安全攻击、安全机理和安全服务，以及这些范畴之间的关系。
被动安全威胁和主动安全威胁之间的差别是什么？被动威胁必须与窃听、或监控、传输发生关系。
电子邮件、文件的传送以及用户/服务器的交流都是可进行监控的传输的例子。
主动攻击包括对被传输的数据加以修改，以及试图获得对计算机系统未经授权的访问。
验证：保证通信实体之一，它声称是。
访问控制：防止未经授权使用的资源（即，谁可以拥有对资源的访问，访问在什么条件下可能发生，那些被允许访问的资源做这个服务控制）。
数据保密：保护数据免受未经授权的披露。
数据完整性：保证接收到的数据是完全作为经授权的实体（即包含任何修改，插入，删除或重播）发送。
不可否认性：提供保护反对否认曾参加全部或部分通信通信中所涉及的实体之一。
可用性服务：系统属性或访问和经授权的系统实体的需求，可用的系统资源，根据系统（即系统是可用的，如果它提供服务，根据系统设计，只要用户要求的性能指标它们）。
第二章1.什么是对称密码的本质成分？明文、加密算法、密钥、密文、解密算法。
4.分组密码和流密码的区别是什么？流密码是加密的数字数据流的一个位或一次一个字节。
块密码是明文块被视为一个整体，用来产生一个相同长度的密文块......分组密码每次处理输入的一组分组，相应的输出一组元素。
流密码则是连续地处理输入元素，每次输出一个元素。
6.列出并简要定义基于攻击者所知道信息的密码分析攻击类型。
惟密文攻击：只知道要解密的密文。
这种攻击一般是试遍所有可能的密钥的穷举攻击，如果密钥空间非常大，这种方法就不太实际。
因此攻击者必须依赖于对密文本身的分析，这一般要运用各种统计方法。
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Known plaintext.The analyst may be able to capture one or more plaintext messages as well as their encryptions. With this knowledge, the analyst may be able to deduce the key on the basis of the way in which the known plaintext is transformed.
书本P36
2.1
a.对b的取值是否有限制？解释原因。
没有限制，b只会使得明文加密后的密文字母统一左移或右移，因此如果是单射的，b改变后依然是单射。
注：答案解答得很坑爹，答了等于没答。现解答如下:
b.判定a不能取哪些值。
2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24.当a大于25时，a也不能是使得a mod 26为这些数的值。
书本P23
2.10简要定义Playfair密码
ThePlayfair algorithmis based on the use of a 55 matrix of letters constructed using a keyword. Plaintext is encrypted two letters at a time using this matrix.
第二章
2.1什么是对称密码的本质成分？
Plaintext, encryption algorithm, secret key, ciphertext, decryption algorithm.
明文加密算法密钥密文解密算法
2.2密码算法中两个基本函数式什么？
Permutation and substitution.
2.5
a.第一个字母t对应A，第二个字母h对应B，e对应C，s对应D，依此类推。随后在句子中重复出现的字母则忽略。结果是
密文: SIDKHKDM AF HCRKIABIE SHIMC KD LFEAILA
明文: basilisk to leviathan blake is contact
b.这是一个单表密码，因此容易被破译
假设明文中频率最高的字母为e，次高的字母为t。注意e=4（e排在第4，a排在第0，没有第26），B=1，t=19，U=20；因此可以得到：
1 = (4a+b) mod 26
20 = (19a+b) mod 26
下式减上式可得19 = 15amod 26，通过反复的错误实验，可得a= 3
然后代入第一条式子可得1 = (12 +b) mod 26，然后得出b= 15
2.7a.加密方法是，先把字母从左到右，从上到下填入矩阵中。然后按第一个密钥的编号，先把编号为1的那一列作为下一个矩阵的第一行，随后的编号按上面的方法填入对应的行。最后按第二个密钥的编号一列一列地写出来。
c.分析a可以取那些值，不可以取那些值。并给出理由。
a与26必须没有大于1的公因子。也就是说a与26互素，或者最大公约数为1.为了说明为什么是这样，先注意到要使E(a,p) = E(a,q) (0 ≤p≤q< 26)成立当且仅当26整除a(p–q).
1.假如a与26互素.则26不能整除a(p–q).这是因为不能减小a/26的这部分而且(p–q)小于26. 2.假如a和26有公因子k> 1.则当q=p+m/k≠p时，p–q=-m/k，显然26能整除a(p–q)，从而E(a,p) = E(a,q).
2.4A good glass in the Bishop's hostel in the Devil's seat—twenty-one degrees and thirteen minutes—northeast and by north—main branch seventh limb east side—shoot from the left eye of the death's head— a bee line from the tree through the shot fifty feet out. (from The Gold Bug, by Edgar Allan Poe)
2.6密文其实指的是一本书中某一页的单词。第一项，534是指第534页。第二项，C2是指第二列。剩余的数字是这一列中的单词。名字DOUGLAS和BIRLSTONE显然是那一页没有出现的单词。太基本了！(from The Valley of Fear, by Sir Arthur Conan Doyle)
书本P33
2.13什么是置换密码？
Atransposition cipherinvolves a permutation of the plaintext letters.
书本P33
2.14什么是隐写术？
Steganography involves concealing the existence of a message.
代换和置换P20
2.3用密码进行通信的两个人需要多少密钥？
对称密码只需要一把，非对称密码要两把P20
2.4分组密码和流密码的区别是什么？
Astream cipheris one that encrypts a digital data stream one bit or one byte at a time. Ablock cipheris one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length.
c.最后一句可能不会包含字母表中的所有字母。如果用第一句的话，随后的句子可以继续填补第一句字母的不全。
2.6The cipher refers to the words in the page of a book. The first entry, 534, refers to page 534. The second entry, C2, refers to column two. The remaining numbers are words in that column. The names DOUGLAS and BIRLSTONE are simply words that do not appear on that page. Elementary! (from The Valley of Fear, by Sir Arthur Conan Doyl么？
1.There is the practical problem of making large quantities of random keys. Any heavily used system might require millions of random characters on a regular basis. Supplying truly random characters in this volume is a significant task.
2.Even more daunting is the problem of key distribution and protection. For every message to be sent, a key of equal length is needed by both sender and receiver. Thus, a mammoth key distribution problem exists.
分组密码每次输入的一组元素，相应地输出一组元素。流密码则是连续地处理输入元素，每次输出一个元素。P20
2.5攻击密码的两种一般方法是什么？
Cryptanalysis and brute force.
密码分析和暴力破解
2.6列出并简要定力基于攻击者所知道信息的密码分析攻击类型。
Ciphertext only. One possible attack under these circumstances is the brute-force approach of trying all possible keys. If the key space is very large, this becomes impractical. Thus, the opponent must rely on an analysis of the ciphertext itself, generally applying various statistical tests to it.
惟密文
已知明文
选择明文
2.7无条件安全密码和计算上安全密码的区别是什么？
An encryption scheme isunconditionally secureif the ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much ciphertext is available. An encryption scheme is said to becomputationally secureif: (1) the cost of breaking the cipher exceeds the value of the encrypted information, and (2) the time required to break the cipher exceeds the useful lifetime of the information.