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"网络安全基础应用与标准第6版英文版答案.pdf" 是一本由William Stallings编写的网络安全教材的解答手册。这本教材的全称为"SOLUTIONS MANUAL NETWORK SECURITY ESSENTIALS: APPLICATIONS AND STANDARDS SIXTH EDITION"，版权为2016年的William Stallings和Pearson Education所有。这本解答手册主要用于解答教材中的问题和练习，并提供了一些选择性的解答与学生们分享。这本教材的作者是William Stallings，他是一位享有盛誉的计算机科学家和网络安全专家。他在该领域有着广泛的经验和研究，并且多次为许多安全教材和标准的编写做出了贡献。他的作品通常都以深入且全面的内容而闻名。 "网络安全基础应用与标准第6版英文版答案.pdf" 这本解答手册是为教师和学生提供辅助资源的。它包含了教材中所提问题的详细解答，并且可以帮助学生更好地理解教材中的概念和原理。解答手册中的解答包括了教材中的各个章节和主题的问题和练习。这本解答手册的内容覆盖了网络安全的基础知识与应用。它包含了网络安全的各个方面，如网络攻击与防御、网络通信与加密、访问控制与认证、安全策略与管理等等。通过学习这本教材和使用解答手册，学生可以建立起对网络安全基本概念的深入理解，并具备必要的技能来应对网络安全问题。该解答手册的版权声明明确表示，未经作者书面许可，任何形式的复制或在互联网上的发布都是禁止的。然而，一些选定的解答可以与学生们分享。这意味着教师可以选择合适的解答与学生们共享，以帮助他们更好地理解和掌握教材内容。综上所述，"网络安全基础应用与标准第6版英文版答案.pdf" 是一本由William Stallings编写的网络安全教材解答手册。它提供了教材中问题与练习的详细解答，并帮助教师和学生更好地理解和应用网络安全的基本知识与技能。然而，在复制和发布该解答手册时，需要遵循相关的版权声明和规定。

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-16-

Delta

< < 4

> > 5

Delta

< < 4

> > 5

2.4 To see that the same algorithm with a reversed key order produces the

correct result, consider Figure 2.2, which shows the encryption process

going down the left-hand side and the decryption process going up the

right-hand side for a 16-round algorithm (the result would be the same

for any number of rounds). For clarity, we use the notation LE

and RE

-17-

for data traveling through the encryption algorithm and LD

and RD

for

data traveling through the decryption algorithm. The diagram indicates

that, at every round, the intermediate value of the decryption process is

equal to the corresponding value of the encryption process with the two

halves of the value swapped. To put this another way, let the output of

the ith encryption round be LE

||RE

concatenated with R

). Then the

corresponding input to the (16 – i)th decryption round is RD

||LD

Let us walk through the figure to demonstrate the validity of the

preceding assertions. To simplify the diagram, it is unwrapped, not

showing the swap that occurs at the end of each iteration. But note that

the intermediate result at the end of the ith stage of the encryption

process is the 2w-bit quantity formed by concatenating LE

and RE

, and

that the intermediate result at the end of the ith stage of the decryption

process is the 2w-bit quantity formed by concatenating LD

and RD

After the last iteration of the encryption process, the two halves of the

output are swapped, so that the ciphertext is RE

||LE

. The output of

that round is the ciphertext. Now take that ciphertext and use it as input

to the same algorithm. The input to the first round is RE

||LE

, which

is equal to the 32-bit swap of the output of the sixteenth round of the

encryption process.

Now we would like to show that the output of the first round of the

decryption process is equal to a 32-bit swap of the input to the

sixteenth round of the encryption process. First, consider the encryption

process. We see that:

= RE

= LE

⊕ F(RE

, K

)

On the decryption side:

= RD

= LE

= RE

= LD

⊕ F(RD

, K

)

= RE

⊕ F(RE

, K

)

= [LE

⊕ F(RE

)] ⊕ F(RE

, K

)

The XOR has the following properties:

[A ⊕ B] ⊕ C = A ⊕ [B ⊕ C]

D ⊕ D = 0

E ⊕ 0 = E

Thus, we have LD

= RE

and RD

= LE

. Therefore, the output of

the first round of the decryption process is LE

||RE

, which is the 32-

bit swap of the input to the sixteenth round of the encryption. This

-20-

€

≈ 2

messages are sent, we expect the same v, and hence

the same key stream, to be used more than once.

d. The key k should be changed sometime before 2

messages are

sent.

2.12 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

is 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.

2.13 In CBC encryption, the input block to each forward cipher operation

(except the first) depends on the result of the previous forward

cipher operation, so the forward cipher operations cannot be

performed in parallel. In CBC decryption, however, the input blocks

for the inverse cipher function (i.e., the ciphertext blocks) are

immediately available, so that multiple inverse cipher operations can

be performed in parallel.

2.14 If an error occurs in transmission of ciphertext block C

, then this

error propagates to the recovered plaintext blocks P

and P

i+1

2.15 After decryption, the last byte of the last block is used to determine

the amount of padding that must be stripped off. Therefore there

must be at least one byte of padding.

2.16 a. Assume that the last block of plaintext is only L bytes long, where L

< 2w/8. The encryption sequence is as follows (The description in

RFC 2040 has an error; the description here is correct.):

1. Encrypt the first (N – 2) blocks using the traditional CBC

technique.

2. XOR P

N–1

with the previous ciphertext block C

N–2

to create Y

N–

3. Encrypt Y

N–1

to create E

N–1

4. Select the first L bytes of E

N–1

to create C

5. Pad P

with zeros at the end and exclusive-OR with E

N–1

create Y

6. Encrypt Y

to create C

N–1

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