密碼編碼學與網(wǎng)絡安全安全第四版 第二章答案翻譯

1、PAGE9第二章什么是對稱密碼的本質(zhì)成分Plaintext, encryption algorithm, secret key, ciphertext, decryption algorithm.明文 加密算法 密鑰 密文 解密算法 密碼算法中兩個基本函數(shù)式什么Permutation and substitution.代換和置換P20用密碼進行通信的兩個人需要多少密鑰對稱密碼只需要一把，非對稱密碼要兩把P20 分組密碼和流密碼的區(qū)別是什么A stream cipher is one that encrypts a digital data stream one bit or one byte

2、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.分組密碼每次輸入的一組元素，相應地輸出一組元素。流密碼則是連續(xù)地處理輸入元素，每次輸出一個元素。P20攻擊密碼的兩種一般方法是什么Cryptanalysis and brute force.密碼分析和暴力破解列出并簡要定力基于攻擊者所知道信息的密碼分析攻擊類型。Ciphertext only. One possible

3、 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.Known analyst may be able to c

4、apture 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 messages to encrypt, the analyst may deliberat

5、ely pick patterns that can be expected to reveal the structure of the key.惟密文已知明文選擇明文無條件安全密碼和計算上安全密碼的區(qū)別是什么An encryption scheme is unconditionally secure if the ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much

6、 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 the information.書本P21簡要定義Caesar密碼The Caesar cipher

7、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.書本P22簡要定義單表代換密碼A monoalphabetic substitution cipher maps a plaintext alphabet to a ciphertext alphabet, so that each letter of the plaintext alphabet maps to a

8、single unique letter of the ciphertext alphabet.書本P23簡要定義Playfair密碼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.書本P26單表代換密碼和奪標代換密碼的區(qū)別是什么A polyalphabetic substitution cipher uses a sepa

9、rate monoalphabetic substitution cipher for each successive letter of plaintext, depending on a key.書本P30一次一密的兩個問題是什么1. 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. Supplying truly ra

10、ndom 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.書本P33什么是置換密碼A transposition cipher

11、involves a permutation of the plaintext letters. 書本P33什么是隱寫術Steganography involves concealing the existence of a message.書本P36習題a.對b的取值是否有限制解釋原因。沒有限制，b只會使得明文加密后的密文字母統(tǒng)一左移或右移，因此如果是單射的，b改變后依然是單射。注：答案解答得很坑爹，答了等于沒答?，F(xiàn)解答如下:b.判定a不能取哪些值。 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24. 當a大于25時，a也不能是使得a mod 26

12、為這些數(shù)的值。c.分析a可以取那些值，不可以取那些值。并給出理由。a與26必須沒有大于1的公因子。也就是說a與26互素，或者最大公約數(shù)為1.為了說明為什么是這樣，先注意到要使E(a, p) = E(a, q) (0 p q 1.則當q = p + m/k p時，p q= -m/k，顯然26能整除a(p q)，從而E(a, p) = E(a, q).有多少種仿射Caesar密碼a有12種可能的值（2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24），b有26種可能的值（0到25），因此總共有12 26 = 312種仿射Caesar密碼。用仿射Caesar

13、密碼加密得到一份密文。頻率最高的字母為B，次高的字母為U，請破譯該密碼。假設明文中頻率最高的字母為e，次高的字母為t。注意e=4（e排在第4，a排在第0，沒有第26），B=1，t=19，U=20；因此可以得到： 1 = (4a + b) mod 2620 = (19a + b) mod 26下式減上式可得19 = 15a mod 26，通過反復的錯誤實驗，可得a = 3然后代入第一條式子可得1 = (12 + b) mod 26，然后得出b = 15 A good glass in the Bishops hostel in the Devils seattwenty-one degrees

14、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)a.第一個字母t對應A，第二個字母h對應B，e對應C，s對應D，依此類推。隨后在句子中重復出現(xiàn)的字母則忽略。結果是密文: SIDKHKDM AF HCRKIABIE S

15、HIMC KD LFEAILA明文: basilisk to leviathan blake is contactb.這是一個單表密碼，因此容易被破譯c.最后一句可能不會包含字母表中的所有字母。如果用第一句的話，隨后的句子可以繼續(xù)填補第一句字母的不全。The 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 words in

16、 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) 密文其實指的是一本書中某一頁的單詞。第一項，534是指第534頁。第二項，C2是指第二列。剩余的數(shù)字是這一列中的單詞。名字DOUGLAS和BIRLSTONE顯然是那一頁沒有出現(xiàn)的單詞。太基本了！(from The Valley of Fear, by Sir Arthur Co

17、nan Doyle)a.加密方法是，先把字母從左到右，從上到下填入矩陣中。然后按第一個密鑰的編號，先把編號為1的那一列作為下一個矩陣的第一行，隨后的編號按上面的方法填入對應的行。最后按第二個密鑰的編號一列一列地寫出來。28107963145CRYPTOGAHIBEATTHETHIRDPILLARFROMTHELEFTOUTSIDETHELYCEUMTHEATRETONIGHTATSEVENIFYOUAREDISTRUSTFULBRINGTWOFRIENDS42810563719NETWORKSCUTRFHEHFTINBROUYRTUSTEAETHGISREHFTEATYRNDIROLTAOUG

18、SHLLETINIBITIHIUOVEUFEDMTCESATWTLEDMNEDLRAPTSETERFOISRNG BUTLF RRAFR LIDLP FTIYO NVSEE TBEHI HTETAEYHAT TUCME HRGTA IOENT TUSRU IEADR FOETO LHMETNTEDS IFWRO HUTEL EITDSb.解密當然是把矩陣倒轉(zhuǎn)順序來用，先用第二個矩陣。首先，根據(jù)第二個密鑰的編碼，把密文按編碼的順序填入列中。然后，再用第一個密鑰，從左到右，從上到下地讀第二個矩陣，按照第一個密鑰的編碼順序，把第一行的字母填入編碼為1的列中，隨后的行依此類推。最后，第一個矩陣從左到右

19、從上到下讀就是明文了。c.雖然這是一個弱的加密方法，但是當加密的內(nèi)容是實時信息，或者企圖竊聽者沒有快速得到好的密碼分析方法（例如，戰(zhàn)略使用），也是可以使用的。加上它除了紙和筆外不需要跟多的工具，而且非常容易記住。SPUTNIKPT BOAT ONE OWE NINE LOST IN ACTION IN BLACKETT STRAIT TWO MILES SW MERESU COVE X CREW OF TWELVE X REQUEST ANY INFORMATIONa.LARGESTBCDFHI/JKMNOPQUVWXYZb.OCURENABDFGHI/JKLMPQSTVWXYZa.UZTBD

20、LGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZb.UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZ c.輪換對稱的行或者列會導致等價的結果。在這一例子中，通過一步的列輪換和三步的行輪換，此問題的a部分的矩陣可以從問題的矩陣得到。a.25! 284b.對于給定的任意5x5配置的矩陣，對于每一行，都有四個輪換變換（循環(huán)右移一格、兩格、三格、四格）是等價的，總共就有五個是一樣的。這五行之中，各自每一行也有這樣的等價關系。所以每個配置就代表著25個等價的配置。因此，總共的密鑰數(shù)量應為25!/25 = 24! 一個混合的Ceasar密碼。移位的數(shù)量由密鑰決

21、定，密鑰決定矩陣中字母的代換。a.Difficulties are things that show what men are.b.Irrationally held truths may be more harmful than reasoned errors.a.我們需要偶數(shù)個字母，因此在最后添加一個”q”。然后按照字母表的位置變換字母。Meetmeattheusual1355201351202085211921112Placeattenrather161213512020514181208518Thaneightoclockq208114597820153121531117必須一次對兩個字母進行計算，第一對是密文的頭兩個字符是字母表的第7和22個位置，對應GV完整的密文：GVUIGVKODZYPUHEKJHUZWFZFWSJSDZMUDZMYCJQMFWWUQRKRb.我們首先求矩陣的逆。注意到加密矩陣的模是(9 7) (4 5) = 43.使用書中求逆矩陣的公式：這里我們利用了等式(43)1 = 23 mod 26。一旦逆矩陣確定，解密就能夠順利執(zhí)行了。來源: LEWA00.考慮矩陣K，由元素kij表示，矩陣Kj是矩陣K的第j列。 EMBED Equation