密码学与网络安全：原理实战与资源库

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《密码编码学与网络安全——原理与实践(第三版)》是一本深入探讨密码学与网络安全领域的重要教材，其课后答案旨在帮助读者巩固课堂所学，理解和掌握相关理论知识。该书覆盖了密码学的基础概念，如加密算法（如欧密会Eurocrypt、美密会Crypto、亚密会Asiacrypt等高级加密会议中的研究成果）、密码分析技术、网络安全策略和实践应用。书中详细介绍了各种密码体制，如对称加密、非对称加密、哈希函数、公钥基础设施（PKI）以及数字签名等，这些都是构建安全通信和保护数据隐私的关键组成部分。此外，它还涉及到了现代密码学中的热门话题，如量子计算对传统密码学的挑战，以及如何设计对抗量子攻击的新型密码方案。对于网络安全部分，课程内容可能涵盖防火墙、入侵检测系统、网络协议的安全分析，以及网络安全管理与策略制定等方面。通过解答课后习题，学生可以锻炼解决实际问题的能力，理解如何在实际环境中实施和维护网络安全措施。该书的课后答案提供了大量的实例和案例分析，以便读者将理论知识应用于实践，并且随着密码学领域的不断进步，这些答案可能会包含最新的研究成果和技术动态。此外，数缘社区作为一个活跃的学术交流平台，不仅提供丰富的电子书库供学习者查阅，还拥有专门的论文库，涵盖了多个国际密码学会议的论文，为研究者提供了丰富的研究资源。参加数学综合讨论区，参与者可以与来自山东大学数学院的优秀研究生互动，共同探讨数学难题，特别是信息安全领域的问题，这对于深入学习和提高技能非常有帮助。同时，社区还提供网络公式编辑器，方便用户在论坛上展示和分享复杂的数学表达式。《密码编码学与网络安全——原理与实践(第三版)》课后答案是密码学和网络安全学习者的宝贵参考资料，它结合了理论与实践，有助于提升读者在这个快速发展的领域的专业素养。而数缘社区作为一个全方位的学习支持平台，无疑为持续学习和研究提供了强大的支撑。

-12-

3.9 The reasoning for the Feistel cipher, as shown in Figure 3.6 applies in the case of

DES. We only have to show the effect of the IP and IP

–1

functions. For encryption,

the input to the final IP

–1

is RE

|| LE

. The output of that stage is the ciphertext.

On decryption, the first step is to take the ciphertext and pass it through IP. Because

IP is the inverse of IP

–1

, the result of this operation is just RE

|| LE

, which is

equivalent to LD

|| RD

. Then, we follow the same reasoning as with the Feistel

cipher to reach a point where LE

= RD

and RE

= LD

. Decryption is completed

by passing LD

|| RD

through IP

–1

. Again, because IP is the inverse of IP

–1

passing the plaintext through IP as the first step of encryption yields LD

|| RD

thus showing that decryption is the inverse of encryption.

3.10 a. Let us work this from the inside out.

|| R

) = L

|| R

) = R

|| L

IP [IP

–1

|| L

)] = R

|| L

) = R

|| L

b. T

|| R

) = L

|| R

IP [IP

–1

|| R

)] = L

|| R

|| L

) = R

|| L

⊕ f(R

, K

)

≠ L

|| R

3.11 PC-1 is essentially the same as IP with every eighth bit eliminated. This would

enable a similar type of implementation. Beyond that, there does not appear to be

any particular cryptographic significance.

3.12

Round number

Bits rotated

3.13 a. The equality in the hint can be shown by listing all 1-bit possibilities:

A ⊕ B

(A ⊕ B)'

A' ⊕ B

We also need the equality A ⊕ B = A' ⊕ B', which is easily seen to be true. Now,

consider the two XOR operations in Figure 3.8. If the plaintext and key for an

encryption are complemented, then the inputs to the first XOR are also

complemented. The output, then, is the same as for the uncomplemented

inputs. Further down, we see that only one of the two inputs to the second

XOR is complemented, therefore, the output is the complement of the output

that would be generated by uncomplemented inputs.

-13-

b. In a chosen plaintext attack, if for chosen plaintext X, the analyst can obtain Y

= E

[X] and Y

= E

[X'], then an exhaustive key search requires only 2

rather

than 2

encryptions. To see this, note that (Y

)' = E

[X]. Now, pick a test value

of the key T and perform E

[X]. If the result is Y

, then we know that T is the

correct key. If the result is (Y

)', then we know that T' is the correct key. If

neither result appears, then we have eliminated two possible keys with one

encryption.

3.14 The result can be demonstrated by tracing through the way in which the bits are

used. An easy, but not necessary, way to see this is to number the 64 bits of the key

as follows (read each vertical column of 2 digits as a number):

2113355-1025554-0214434-1123334-0012343-2021453-0202435-0110454-

1031975-1176107-2423401-7632789-7452553-0858846-6836043-9495226-

The first bit of the key is identified as 21, the second as 10, the third as 13, and so on.

The eight bits that are not used in the calculation are unnumbered. The numbers 01

through 28 and 30 through 57 are used. The reason for this assignment is to clarify

the way in which the subkeys are chosen. With this assignment, the subkey for the

first iteration contains 48 bits, 01 through 24 and 30 through 53, in their natural

numerical order. It is easy at this point to see that the first 24 bits of each subkey

will always be from the bits designated 01 through 28, and the second 24 bits of

each subkey will always be from the bits designated 30 through 57.

3.15 a. No. For example, suppose C

is corrupted. The output block P

depends only

on the input blocks C

and C

b. An error in P

affects C

. But since C

is input to the calculation of C

, C

affected. This effect carries through indefinitely, so that all ciphertext blocks are

affected. However, at the receiving end, the decryption algorithm restores the

correct plaintext for blocks except the one in error. You can show this by

writing out the equations for the decryption. Therefore, the error only effects

the corresponding decrypted plaintext block.

3.16 Nine plaintext characters are affected. The plaintext character corresponding to

the ciphertext character is obviously altered. In addition, the altered ciphertext

character enters the shift register and is not removed until the next eight characters

are processed.

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