密码编码学与网络安全中文答案
【篇一：密码编码学与网络安全第四版 第二章答案翻译】
是对称密码的本质成分？
plaintext, encryption algorithm, cret 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？
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 ud 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。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 the 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 itlf, 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.
chon plaintext. if the analyst is able to choo 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 cure 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 cure 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 uful lifetime of the information.
书本p21
2.8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p22
2.9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p23
2.10简要定义playfair密码
the playfair algorithm is bad on the u of a 5?5 matrix of letters constructed using a keyword. plaintext is encrypted two letters at a time using this matrix.
书本p26
2.11单表代换密码和夺标代换密码的区别是什么？
a polyalphabetic substitution cipher us a parate monoalphabetic substitution cipher for each successive letter of plaintext, depending on a key.
书本p30
2.12一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 ud
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 nt, a key of equal length is needed by both nder and receiver. thus, a mammoth key distribution problem exists.
书本p33
2.13什么是置换密码？
a transposition cipher involves 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.
书本p36
2.7.3习题
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:
若要e?k,p??e?k,q?，则ap?b??aq?b?mod26不成立，则ap?aq?0mod26不成立。这时b已经消掉了，因此显然b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).
注：由a.知a?p?q??0mod26不成立，则26不能整除a?p?q?.由文中暗示0?p?26,0?q?26，这个其实开头我也不知道，后来觉得应该是这样。
因为 p与q不相等，所以p?q?26，因此p?q不可能是26的整数倍，但是p?q有可能是2或13的整数倍。因此a不能是2或13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
