密码编码学与网络安全第四版-第二章答案翻译(共9页).doc

《密码编码学与网络安全第四版-第二章答案翻译(共9页).doc》由会员分享，可在线阅读，更多相关《密码编码学与网络安全第四版-第二章答案翻译(共9页).doc（9页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、精选优质文档-倾情为你奉上第二章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 str

2、eam 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 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

3、型。Ciphertext 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

4、.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.Chosen plaintext. If the analyst is able to choose the m

5、essages 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 uniquel

6、y 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 th

7、e 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 alphabet, s

8、o 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 55 matrix of letters constructed using a keyword. Plaintext is encrypted two letters at a time using this matrix.书本P262.11单表代换

9、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 might require

10、 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. Thus, a mammot

11、h 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。注：答案解答得很坑爹，答了等于没答。现解

12、答如下: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 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).2.2有多少种仿射Caesar密码？a有12种可能的值（2, 4, 6,

13、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 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a = 3然后代入第一条式子可得1 = (12

14、+ b) mod 26，然后得出b = 15 2.4 A good glass in the Bishops hostel in the Devils seattwenty-one degrees and thirteen minutesnortheast and by northmain branch seventh limb east sideshoot from the left eye of the deaths head a bee line from the tree through the shot fifty feet out. (from The Gold Bug, by E

15、dgar Allan Poe)2.5a.第一个字母t对应A，第二个字母h对应B，e对应C，s对应D，依此类推。随后在句子中重复出现的字母则忽略。结果是密文: SIDKHKDM AF HCRKIABIE SHIMC KD LFEAILA明文: basilisk to leviathan blake is contactb.这是一个单表密码，因此容易被破译c.最后一句可能不会包含字母表中的所有字母。如果用第一句的话，随后的句子可以继续填补第一句字母的不全。2.6The cipher refers to the words in the page of a book. The first entry

16、, 534, refers to page 534. The second entry, C2, refers to column two. The remaining numbers are words in that column. The names DOUGLAS and BIRLSTONE are simply words that do not appear on that page. Elementary! (from The Valley of Fear, by Sir Arthur Conan Doyle)2.6 密文其实指的是一本书中某一页的单词。第一项，534是指第534

17、页。第二项，C2是指第二列。剩余的数字是这一列中的单词。名字DOUGLAS和BIRLSTONE显然是那一页没有出现的单词。太基本了！(from The Valley of Fear, by Sir Arthur Conan Doyle)2.7a.加密方法是，先把字母从左到右，从上到下填入矩阵中。然后按第一个密钥的编号，先把编号为1的那一列作为下一个矩阵的第一行，随后的编号按上面的方法填入对应的行。最后按第二个密钥的编号一列一列地写出来。28107963145CRYPTOGAHIBEATTHETHIRDPILLARFROMTHELEFTOUTSIDETHELYCEUMTHEATRETONIGHT

18、ATSEVENIFYOUAREDISTRUSTFULBRINGTWOFRIENDS42810563719NETWORKSCUTRFHEHFTINBROUYRTUSTEAETHGISREHFTEATYRNDIROLTAOUGSHLLETINIBITIHIUOVEUFEDMTCESATWTLEDMNEDLRAPTSETERFOISRNG BUTLF RRAFR LIDLP FTIYO NVSEE TBEHI HTETAEYHAT TUCME HRGTA IOENT TUSRU IEADR FOETO LHMETNTEDS IFWRO HUTEL EITDSb.解密当然是把矩阵倒转顺序来用，先用第二

19、个矩阵。首先，根据第二个密钥的编码，把密文按编码的顺序填入列中。然后，再用第一个密钥，从左到右，从上到下地读第二个矩阵，按照第一个密钥的编码顺序，把第一行的字母填入编码为1的列中，随后的行依此类推。最后，第一个矩阵从左到右从上到下读就是明文了。c.虽然这是一个弱的加密方法，但是当加密的内容是实时信息，或者企图窃听者没有快速得到好的密码分析方法（例如，战略使用），也是可以使用的。加上它除了纸和笔外不需要跟多的工具，而且非常容易记住。2.8SPUTNIK2.9PT BOAT ONE OWE NINE LOST IN ACTION IN BLACKETT STRAIT TWO MILES SW ME

20、RESU COVE X CREW OF TWELVE X REQUEST ANY INFORMATION2.10a.LARGESTBCDFHI/JKMNOPQUVWXYZb.OCURENABDFGHI/JKLMPQSTVWXYZ2.11a.UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZb.UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZ c.轮换对称的行或者列会导致等价的结果。在这一例子中，通过一步的列轮换和三步的行轮换，此问题的a部分的矩阵可以从问题2.10a的矩阵得到。2.12a.25! 284b.对于给定的任意5x5配置的矩阵，对

21、于每一行，都有四个轮换变换（循环右移一格、两格、三格、四格）是等价的，总共就有五个是一样的。这五行之中，各自每一行也有这样的等价关系。所以每个配置就代表着25个等价的配置。因此，总共的密钥数量应为25!/25 = 24!2.13 一个混合的Ceasar密码。移位的数量由密钥决定，密钥决定矩阵中字母的代换。2.14a.Difficulties are things that show what men are.b.Irrationally held truths may be more harmful than reasoned errors.2.15a.我们需要偶数个字母，因此在最后添加一个”

22、q”。然后按照字母表的位置变换字母。Meetmeattheusual1355201351202085211921112Placeattenrather161213512020514181208518Thaneightoclockq208114597820153121531117必须一次对两个字母进行计算，第一对是密文的头两个字符是字母表的第7和22个位置，对应GV完整的密文：GVUIGVKODZYPUHEKJHUZWFZFWSJSDZMUDZMYCJQMFWWUQRKRb.我们首先求矩阵的逆。注意到加密矩阵的模是(9 7) (4 5) = 43.使用书中求逆矩阵的公式：这里我们利用了等式(43

23、)1 = 23 mod 26。一旦逆矩阵确定，解密就能够顺利执行了。来源: LEWA00.2.16考虑矩阵K，由元素kij表示，矩阵Kj是矩阵K的第j列。and通过选择下列n组明文就能暴露出K矩阵的每一列:(B, A, A, , A, A) K1(A, B, A, , A, A) K2:(A, A, A, , A, B) Kn注意，A代表0，B代表12.172.17a.7 134b.7 134c.134d.10 134e.24 132f.24 (132 1) 13g.37648h.23530i.2.18密钥:legleglegle明文:explanation密文:PBVWETLXOZR2.19

24、a.sendmoremoney18413312141741214134249017231521141111289141410931218232515127BECKJDMSXZPMHb.cashnotneeded201871314191344343254223221519519211284141410931218232515127BECKJDMSXZPMH2.20your package ready Friday 21st room three Please destroy this immediately.2.21a.把信息填入8列的矩阵总，每个整数告诉你应选择各行的第几个字母。结果:He sitteth between the cherubims. The isles may be glad thereof. As the rivers in the south.b.挺安全的，每一行是1/8的概率。因此如果密文长度是8的倍数，那么明文也可能是8的倍数。c.不是很安全，Lord Peter就破解了它。(出自 The Nine Tailors)专心-专注-专业