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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。
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。
现代密码学第四版答案第一章简介1.1 密码学概述1.1.1 什么是密码学？密码学是研究通信安全和数据保护的科学和艺术。
它涉及使用各种技术和方法来保护信息的机密性、完整性和可用性。
1.1.2 密码学的分类密码学可以分为两个主要方向：对称密码学和非对称密码学。
•对称密码学：在对称密码学中，发送者和接收者使用相同的密钥来进行加密和解密。
•非对称密码学：在非对称密码学中，发送者和接收者使用不同的密钥来进行加密和解密。
1.2 密码系统的要素1.2.1 明文和密文•明文（plaintext）：未经加密的原始消息。
•密文（ciphertext）：经过加密后的消息。
1.2.2 密钥密钥是密码系统的核心组成部分，它用于加密明文以生成密文，或者用于解密密文以恢复明文。
密钥应该是保密的，只有合法的用户才能知道密钥。
1.2.3 加密算法加密算法是用来将明文转换为密文的算法。
加密算法必须是可逆的，这意味着可以使用相同的密钥进行解密。
1.2.4 加密模式加密模式是规定了加密算法如何应用于消息的规则。
常见的加密模式包括电子密码本（ECB）、密码块链路（CBC）和计数器模式（CTR）等。
1.3 密码的安全性密码的安全性取决于密钥的长度、加密算法的复杂度以及密码系统的安全性设计。
第二章对称密码学2.1 凯撒密码凯撒密码是一种最早的加密方式，它将字母按照给定的偏移量进行位移。
例如，偏移量为1时，字母A加密后变为B，字母B变为C，以此类推。
2.2 DES加密算法DES（Data Encryption Standard）是一种对称密码算法，它使用56位密钥对64位的明文进行加密。
DES算法包括初始置换、16轮迭代和最终置换三个阶段。
2.3 AES加密算法AES（Advanced Encryption Standard）是一种对称密码算法，它使用128位、192位或256位的密钥对128位的明文进行加密。
AES算法使用了替代、置换和混淆等操作来保证对抗各种密码攻击。
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。
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:()()() ,,mod 260mod 26E k p E k q ap b aq b ap aq b b ≠+≡+-≡若要，则不成立，则不成立。
这时已经消掉了，因此显然不会影响算法的映射特性。
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。
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 ).()() .0mod 2626.26,026 2626213213a a p q a p q p q p q p q p q p q a -≡-≤≤≤≤-<--注：由知不成立，则不能整除由文中暗示0，这个其实开头我也不知道，后来觉得应该是这样。
因为与不相等，所以，因此不可能是的整数倍，但是有可能是或的整数倍。
因此不能是或的整数倍。
2.2有多少种仿射Caesar 密码？a 有12种可能的值（2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24），b 有26种可能的值（0到25），因此总共有12 ⨯ 26 = 312种仿射Caesar 密码。
2.3用仿射Caesar 密码加密得到一份密文。
频率最高的字母为B ，次高的字母为U ，请破译该密码。
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 = 15a mod 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()()()()()()() gcd 15,26115261261511151141124343141124115111121511115261526152152615115s t s t =+==+=+=⨯+=+=-⨯+-=-⨯-+--=--⨯--+⎡⎤⎣⎦注意：答案说经过反复错误试验，这个解答很坑爹，实际上是可以精确算出的。
