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1、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
2、 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。Ci
3、phertext 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.Kn
4、own 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 mes
5、sages 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 unique
6、ly 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 t
7、he 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 alphab
8、et, 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 P2
9、62.11 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 used system m
10、ight 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. Th
11、us, 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
12、。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 ,贝Uap b aq b mod26不成立，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
13、) = EG，q) (0医Wq 1则当q = p + m/k wp时，p -q= -m/k,显然26能整除a(p - q),从而 E(a, p) = EG, q).注：由a知a p q0mod26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1田勺整数倍。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 )
14、, 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to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：I = (4a + b) mod 2620 = (19a + b) mod 26下式减上式可得19 = 15a mod 26,通过反复的错误实验，可得然后代入第一条式子可得1 = (12 + b) mod 26 ,然后得出b = 15注意：答案说经过反
15、复错误试验，这个解答很坑爹，实际上是可以精确算 出的。因为gcd 15,261,因此由裴蜀定理，必存在整数s和t,使得15s 26t 1。由欧几里得算法26 15 1115 11 4II 2 4 34 3 1因此 4112 41,因此 15 1111 2 15 111因此 1526 1526 1521526 151最后得出15 7 26 4 1所以 15 7 1mod 26119所以一 7 mod 26,从而a 19 7 133 3mod 26,3就是这样的出来的.1515于是 p C b a 1 mod 26, W p C 15 3 1 mod 26,显然 31 9mod 262.4 A good glass in the Bishops hostel in the Devils seattwenty-one degrees and thirteen minutes northeast and by north main branch seventh limb east side shoot from the left eye of the deaths head a bee line from the tree throu
