密码编码学与网络安全英文答案优质资料.doc

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1、密码编码学与网络安全英文答案优质资料(可以直接使用，可编辑 优质资料，欢迎下载）Chapter 1:Introduction5Chapter 2:Classical Encryption Techniques7Chapter 3:Block Ciphers and the Date Encryption Standard13Chapter 4:Finite Fields21Chapter 5:Advanced Encryption Standard28Chapter 6:More on Symmetric Ciphers33Chapter 7:Confidentiality Using Sym

2、metric Encryption38Chapter 8:Introduction to Number Theory42Chapter 9:Public-Key Cryptography and RSA46Chapter 10:Key Management; Other Public-Key Cryptosystems55Chapter 11:Message Authentication and Hash Functions59Chapter 12:Hash and MAC Algorithms62Chapter 13:Digital Signatures and Authentication

3、 Protocols66Chapter 14:Authentication Applications71Chapter 15:Electronic Mail Security73Chapter 16:IP Security76Chapter 17:Web Security80Chapter 18:Intruders83Chapter 19:Malicious Software87Chapter 20:Firewalls89Answers to Questions1.1The OSI Security Architecture is a framework that provides a sys

4、tematic way of defining the requirements for security and characterizing the approaches to satisfying those requirements. The document defines security attacks, mechanisms, and services, and the relationships among these categories.1.2Passive attacks have to do with eavesdropping on, or monitoring,

5、transmissions. Electronic mail, file transfers, and client/server exchanges are examples of transmissions that can be monitored. Active attacks include the modification of transmitted data and attempts to gain unauthorized access to computer systems.1.3Passive attacks: release of message contents an

6、d traffic analysis. Active attacks: masquerade, replay, modification of messages, and denial of service.1.4Authentication: The assurance that the communicating entity is the one that it claims to be. Access control: The prevention of unauthorized use of a resource (i.e., this service controls who ca

7、n have access to a resource, under what conditions access can occur, and what those accessing the resource are allowed to do). Data confidentiality: The protection of data from unauthorized disclosure. Data integrity: The assurance that data received are exactly as sent by an authorized entity (i.e.

8、, contain no modification, insertion, deletion, or replay). Nonrepudiation: Provides protection against denial by one of the entities involved in a communication of having participated in all or part of the communication. Availability service: The property of a system or a system resource being acce

9、ssible and usable upon demand by an authorized system entity, according to performance specifications for the system (i.e., a system is available if it provides services according to the system design whenever users request them). 1.5See Table 1.3.Answers toProblems1.1Release of message contentsTraf

10、fic analysisMasqueradeReplayModification of messagesDenial of servicePeer entity authenticationYData origin authenticationYAccess controlYConfidentialityYTraffic flow confidentialityYData integrityYYNon-repudiationYAvailabilityY1.2Release of message contentsTraffic analysisMasqueradeReplayModificati

11、on of messagesDenial of serviceEnciphermentYDigital signatureYYYAccess controlYYYYYData integrityYYAuthentication exchangeYYYYTraffic paddingYRouting controlYYYNotarizationYYYChapter 2Classical Encryption TechniquesrAnswers to Questions2.1Plaintext, encryption algorithm, secret key, ciphertext, decr

12、yption algorithm.2.2Permutation and substitution.2.3One key for symmetric ciphers, two keys for asymmetric ciphers.2.4A 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 used to produc

13、e a ciphertext block of equal length.2.5Cryptanalysis and brute force.2.6Ciphertext 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 th

14、e ciphertext itself, 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 plaintex

15、t is transformed. Chosen plaintext. If the analyst is able to choose the messages to encrypt, the analyst may deliberately pick patterns that can be expected to reveal the structure of the key.2.7An encryption scheme is unconditionally secure if the ciphertext generated by the scheme does not contai

16、n enough information to determine uniquely 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

17、 cipher exceeds the useful lifetime of the information.2.8The 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 25.2.9A monoalphabetic substitution cipher maps a plaintext alphabet to a ciphertext al

18、phabet, so that each letter of the plaintext alphabet maps to a single unique letter of the ciphertext alphabet.2.10The 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.2.11A polyalphabetic s

19、ubstitution cipher uses a separate monoalphabetic substitution cipher for each successive letter of plaintext, depending on a key.2.121. There is the practical problem of making large quantities of random keys. Any heavily used system might require millions of random characters on a regular basis. S

20、upplying 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 mammoth key distribution problem exists.2.13A transpositio

21、n cipher involves a permutation of the plaintext letters.2.14Steganography involves concealing the existence of a message.Answers to Problems2.1a.No. A change in the value of b shifts the relationship between plaintext letters and ciphertext letters to the left or right uniformly, so that if the map

22、ping is one-to-one it remains one-to-one.不，在b值变化之间的关系和明文字母转换密文字母到左或右一致，所以，如果映射是一对一的 它仍然是一对一的。b.2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24. Any value of a larger than 25 is equivalent to a mod 26.任何一个大于25的值是相当于一个mod 26。c.The values of a and 26 must have no common positive integer factor other tha

23、n 1. This is equivalent to saying that a and 26 are relatively prime, or that the greatest common divisor of a and 26 is 1. To see this, first note that E(a, p) = E(a, q) (0 p q 1. Then E(a, p) = E(a, q), if q = p + m/k p.一个26的值必须没有共同的正整数因子大于1。这等于说，和26互质，或一个和26的最大公约数为1。看到这一点，首先注意E（A，P）= E（a，q）（0Pq26

24、）当且仅当（PQ）能被26整除。1。假设一个和26互质。然后，一个（PQ）是不能被26整除，因为没有办法降低分数/ 26（PQ）小于26。2。假设一个26有一个共同的因子K1。然后，E（A，P）=E（a，q），如果q=P+ M/KP.2.2There are 12 allowable values of a (1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25). There are 26 allowable values of b, from 0 through 25). Thus the total number of distinct affine Cae

25、sar ciphers is 12 26 = 312.有一个12的允许值（1，3，5，7，9，11，15，17，19，21，23，25）。有B 26的允许值，从0到25）。因此，不同的仿射凯撒密码的总数是1226 =312。2.3Assume that the most frequent plaintext letter is e and the second most frequent letter is t. Note that the numerical values are e = 4; B = 1; t = 19; U = 20. Then we have the following

26、 equations:假设最频繁的明文字母是E和第二个最常见的字母T.注意值E = 4；乙= 1；t = 19；U = 20。然后我们有以下方程：1 = (4a + b) mod 2620 = (19a + b) mod 26Thus, 19 = 15a mod 26. By trial and error, we solve: a = 3.因此，19 = 15a mod 26。通过试验和错误，我们解决：=3。Then 1 = (12 + b) mod 26. By observation, b = 15.然后1 =（12 + B）国防部26。通过观察，B=15。2.4A good glass

27、 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 Edgar Allan Poe)一个好的玻璃在主教的宿舍在魔鬼的座二零一度十三分

28、东北偏北的主要分支的第七肢芽从东边的死亡的头-左眼直线从树上通过五十脚射出去。（从黄金的错误，埃德加爱伦坡）2.5a.The first letter t corresponds to A, the second letter h corresponds to B, e is C, s is D, and so on. Second and subsequent occurrences of a letter in the key sentence are ignored. The result第一个字母T对应一个，第二个字母H对应于B，是C，S D，等等。在关键句的一封信，第二，随后出现被忽

29、略。结果ciphertext密文: SIDKHKDM AF HCRKIABIE SHIMC KD LFEAILAplaintext明文: basilisk to leviathan blake is contactb.It is a monalphabetic cipher and so easily breakable.这是一个monalphabetic密码等易碎的。c.The last sentence may not contain all the letters of the alphabet. If the first sentence is used, the second and

30、 subsequent sentences may also be used until all 26 letters are encountered.最后一句话可能不包含所有的字母。如果第一句的使用，第二和随后的句子也可以使用直到遇到所有的26个字母。2.6The cipher refers to the words in the page of a book. The first entry, 534, refers to page 534. The second entry, C2, refers to column two. The remaining numbers are word

31、s 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)密码是指词在一本书的页面。第一项，534，是指534页。第二项，C2，指的是两列。剩下的数字是该列中的话。名称道格拉斯和伯尔斯通只是词没有出现在那一页。小学！（从恐惧的山谷，Sir亚瑟柯南道尔）2.7a.28107963145CRYPTOGAHIBEATTHETHIRDP

32、ILLARFROMTHELEFTOUTSIDETHELYCEUMTHEATRETONIGHTATSEVENIFYOUAREDISTRUSTFULBRINGTWOFRIENDS42810563719NETWORKSCUTRFHEHFTINBROUYRTUSTEAETHGISREHFTEATYRNDIROLTAOUGSHLLETINIBITIHIUOVEUFEDMTCESATWTLEDMNEDLRAPTSETERFOISRNG BUTLF RRAFR LIDLP FTIYO NVSEE TBEHI HTETAEYHAT TUCME HRGTA IOENT TUSRU IEADR FOETO LHM

33、ETNTEDS IFWRO HUTEL EITDSb.The two matrices are used in reverse order. First, the ciphertext is laid out in columns in the second matrix, taking into account the order dictated by the second memory word. Then, the contents of the second matrix are read left to right, top to bottom and laid out in co

34、lumns in the first matrix, taking into account the order dictated by the first memory word. The plaintext is then read left to right, top to bottom.两个矩阵是用相反的顺序。首先，密文是奠定在列中的第二矩阵，考虑到由第二记忆单词的顺序。然后，第二矩阵的内容是从左到右，顶部和底部放置在列中的第一个矩阵，考虑到由第一记忆单词的顺序。明文是从左到右，顶部底部c.Although this is a weak method, it may have use

35、with time-sensitive information and an adversary without immediate access to good cryptanalysis (e.g., tactical use). Plus it doesnt require anything more than paper and pencil, and can be easily remembered.虽然这是一个弱的方法，它可以使用的时间敏感的信息和一个对手没有立即获得良好的密码分析（例如，战术运用）。再加上它不需要任何比纸和铅笔，并可以很容易地记住。2.8SPUTNIK2.9PT

36、BOAT ONE OWE NINE LOST IN ACTION IN BLACKETT STRAIT TWO MILES SW MERESU COVE X CREW OF TWELVE X REQUEST ANY INFORMATION2.10a.LARGESTBCDFHI/JKMNOPQUVWXYZb.OCURENABDFGHI/JKLMPQSTVWXYZ2.11a.UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZb.UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZc.A cyclic rotation of rows and/or

37、columns leads to equivalent substitutions. In this case, the matrix for part a of this problem is obtained from the matrix of Problem 2.10a, by rotating the columns by one step and the rows by three steps.一个循环旋转的行和/或柱的等效替换。在这种情况下，对这个问题的一部分是从问题2.10a矩阵的矩阵，通过一步和三步行旋转柱。2.12a.25! 284b.Given any 5x5 confi

38、guration, any of the four row rotations is equivalent, for a total of five equivalent configurations. For each of these five configurations, any of the four column rotations is equivalent. So each configuration in fact represents 25 equivalent configurations. Thus, the total number of unique keys is

39、 25!/25 = 24!给出任何5x5配置，任何四行的旋转是等效的，一共有五个等效组态。对于这五种配置，任何四列的旋转是等价的。所以事实上每个配置代表25等效组态。因此，独特的键的总数是25！2524。2.13A mixed Caesar cipher. The amount of shift is determined by the keyword, which determines the placement of letters in the matrix.一种混合的恺撒密码。偏移量是由关键字确定，这决定了字母在矩阵中的位置。2.14a.Difficulties are things

40、that show what men are.b.Irrationally held truths may be more harmful than reasoned errors.2.15a.We need an even number of letters, so append a q to the end of the message. Then convert the letters into the corresponding alphabetic positions:我们需要一个偶数的信件，所以添加一个“Q”消息的结束。然后将信件放入相应的字母位置：Meetmeattheusual

41、1355201351202085211921112Placeattenrather161213512020514181208518Thaneightoclockq208114597820153121531117The calculations proceed two letters at a time. The first pair:进行计算的两个字母的时候。第一对：The first two ciphertext characters are alphabetic positions 7 and 22, which correspond to GV. The complete ciphert

42、ext:前两个密文字符是字母的位置7和22，对应于GV。完整的密文：GVUIGVKODZYPUHEKJHUZWFZFWSJSDZMUDZMYCJQMFWWUQRKRb. We first perform a matrix inversion. Note that the determinate of the encryption matrix is (9 7) (4 5) = 43. Using the matrix inversion formula from the book:我们先进行矩阵求逆。5）= 43。7）（4注意加密矩阵的行列式（9利用书中的矩阵求逆公式：Here we used

43、 the fact that (43)1 = 23 in Z26. Once the inverse matrix has been determined, decryption can proceed. Source: LEWA00.2.16Consider the matrix K with elements kij to consist of the set of column vectors Kj, where:andThe ciphertext of the following chosen plaintext n-grams reveals the columns of K:(B, A, A, , A, A) K1(A, B, A, , A, A) K2:(A, A, A, , A, B) Kn2.17a.7 134b.7 134c.134d.10 134e.24 132f.24 (132 1) 13g.37648h.23530i.1572482.18key:legleglegleplaintext:explanationciphertex