密码学与网络安全课后答案解析

4星 · 超过85%的资源需积分: 15 199 浏览量更新于2024-07-27 1 收藏 802KB PDF 举报

"密码学与网络安全课后答案，包含第四版的详细解答，由William Stallings编写，涵盖了多个关键章节的复习问题和作业问题的解决方案。" 在密码学与网络安全领域，理解和掌握各种加密技术和网络防护策略是至关重要的。William Stallings的《密码学与网络安全：原理与实践》第四版是一本广泛使用的教材，它深入浅出地讲解了这个领域的核心概念。课后答案提供了学习者检查理解、巩固知识的重要途径。 1. 第一章：介绍这一章通常会涵盖密码学的基本定义，历史发展，以及它在网络安全中的作用。可能会讨论公钥和私钥加密的区别，以及加密、解密和数字签名的基本概念。 2. 第二章：古典加密技术古典加密方法包括替换密码（如凯撒密码）和置换密码（如维吉尼亚密码）。这些方法虽然简单，但它们为现代加密算法提供了基础。 3. 第三章：分组密码和数据加密标准（DES） DES是早期广泛使用的对称加密算法，它的工作模式和安全性分析在这里会被详细解释，同时可能还会讨论其局限性和被AES取代的原因。 4. 第四章：有限域有限域在公钥密码学中起着核心作用，如RSA算法就需要利用模算术和有限域的性质。这一章会介绍如何操作有限域内的元素，以及它们在密码学中的应用。 5. 第五章：高级加密标准（AES） AES是目前最常用的对称加密标准，它的设计原理、工作流程以及密钥管理会在这一章中详细阐述。 6. 第六章：更多关于对称密码可能会涵盖其他对称加密算法，如三重DES（3DES）和Blowfish，以及它们相对于DES的优势和改进。 7. 第七章：基于对称加密的机密性这一部分将讨论如何使用对称加密来保护数据的机密性，包括密钥管理和安全传输协议如SSL/TLS。 8. 第八章：数论导论数论是公钥密码学的基础，如欧几里得算法、模逆元和大整数分解问题（RSA的安全基础）等主题会在此章节出现。每个章节的课后答案都提供了深入的解析，帮助学生理解和解决实际问题，这对于深入学习密码学原理和实践至关重要。通过解决这些问题，学生可以增强对密码学复杂概念的理解，并能够应用这些知识来解决实际的网络安全问题。

-16-

encryption is embedded in applications or utility functions in such a way as to

preclude a hardware implementation. Accordingly, the speed of execution of the

algorithm becomes a concern. Ease of analysis: Although we would like to make

our algorithm as difficult as possible to cryptanalyze, there is great benefit in

making the algorithm easy to analyze. That is, if the algorithm can be concisely and

clearly explained, it is easier to analyze that algorithm for cryptanalytic

vulnerabilities and therefore develop a higher level of assurance as to its strength.

3.7 The S-box is a substitution function that introduces nonlinearity and adds to the

complexity of the transformation.

3.8 The avalanche effect is a property of any encryption algorithm such that a small

change in either the plaintext or the key produces a significant change in the

ciphertext.

3.9 Differential cryptanalysis is a technique in which chosen plaintexts with particular

XOR difference patterns are encrypted. The difference patterns of the resulting

ciphertext provide information that can be used to determine the encryption key.

Linear cryptanalysis is based on finding linear approximations to describe the

transformations performed in a block cipher.

NSWERS TO

ROBLEMS

3.1 a. For an n-bit block size are 2

possible different plaintext blocks and 2

possible

different ciphertext blocks. For both the plaintext and ciphertext, if we treat the

block as an unsigned integer, the values are in the range 0 through 2

– 1. For a

mapping to be reversible, each plaintext block must map into a unique

ciphertext block. Thus, to enumerate all possible reversible mappings, the block

with value 0 can map into anyone of 2

possible ciphertext blocks. For any

given mapping of the block with value 0, the block with value 1 can map into

any one of 2

– 1 possible ciphertext blocks, and so on. Thus, the total number

of reversible mappings is (2

)!.

b. In theory, the key length could be log

)! bits. For example, assign each

mapping a number, from 1 through (2

)! and maintain a table that shows the

mapping for each such number. Then, the key would only require log

)! bits,

but we would also require this huge table. A more straightforward way to

define the key is to have the key consist of the ciphertext value for each

plaintext block, listed in sequence for plaintext blocks 0 through 2

– 1. This is

what is suggested by Table 3.1. In this case the key size is n

× 2n and the huge

table is not required.

3.2 Because of the key schedule, the round functions used in rounds 9 through 16 are

mirror images of the round functions used in rounds 1 through 8. From this fact

we see that encryption and decryption are identical. We are given a ciphertext c.

剩余106页未读，继续阅读

chwcasiie

粉丝: 0
资源: 1

密码学与网络安全课后答案解析

密码编码学与网络安全课后习题解析

《密码编码学与网络安全》第四版课后答案解析

密码学与网络安全：原理实战与资源库

密码编码学与网络安全课后答案1

中科大研究生密码学-密码编码学和网络安全-课后答案

密码编码学与网络安全(课后习题答案)

密码学第四版课后答案

密码编码学与网络安全_课后习题答案(全).doc

密码编码学与网络安全（第四版） 课后习题答案

密码学原理与实践课后习题答案.7z

最新资源

密码编码学与网络安全（第四版）课后习题答案