交大網(wǎng)絡(luò)與信息安全第二次作業(yè)

1、網(wǎng)絡(luò)與信息安全第二次作業(yè)Assign 2 - InetSec 力口密Assign 2 - InetSec CryptoSecret and Public Key Cryptography密鑰和公鑰的密碼技術(shù)1. How many DES keys, on the average, encrypt a particular plaintext block to a particular ciphertext block? Kaufman §3.3 (10 points)一般來說，把一段詳細(xì)的評(píng)述文字加密成一段加密文字，需要多少位DES密碼？答：DSE 一般采用56位長(zhǎng)度的Key,所以總

2、共有256種可能，這個(gè)數(shù)字 大約是7.2X10的16次方。2. Suppose the DES mangler function mapped every 32-bit value to zero, regardless of the value of its input. Whatfunction would DES then compute?Kaufman3.5 (10points)假設(shè)這種DES切割函數(shù)就是不管輸入什么值每32位的值都映射成0 ,那么DES是什么函數(shù)又是怎么計(jì)算的呢？答：置換函數(shù) Li=Ki-1;Ri=Li+1*F(Ri-1,Ki)3. It is said that t

3、he initial and final permutations of all 64 bits in DES operation do not enhance the cryptographic strengthof the encryption algorithm. Could you provide an explanation without using sophisticated mathematics?(10points)據(jù)說，64位前后交換位置這種 DES方法不能提高加密算法的密碼強(qiáng)度。 不使用復(fù)雜的數(shù)學(xué)理論你能提供一個(gè)解釋來說明嗎？答：因?yàn)檩敵雠帕?（輸入排列）-1,每一個(gè)排列

4、都是一個(gè)有次序的 Bryant-Tree排列，所以并沒有安全上的改善。4. Compute the number of 64-bit encryption operations performed for an n bit plaintext using CBC, k-bit OFB and k-bit CFB. Count all encryption operations, not just operations performed on the plaintext itself. Take as an example, n = 1024 and k = 32.(10 points)計(jì)算6

5、4位的數(shù)字加密操作通過使用CBC , k位OFB和k位CFB把它變成一個(gè)n位的評(píng)述文字。計(jì)算所有的加密操作，這些操作不僅僅運(yùn)行在這 個(gè)評(píng)述文字本身。舉個(gè)例子n=1024 和k =32 。答：密文區(qū)段串接 (Cipher BlockChaining, CBC) 模式k-位元密文反饋 (k-bits Cipher Feedback, CFB)模式加密運(yùn)算程序：SR1 =IV C1 = Fj (EK(SR1) P1 SRm = Sj(SRm-1) | Cm-1 ;m = 2, 3, 4, , N Cm = Fj (EK(SRm) Pm ;m = 2, 3, 4, ,N C=C1 | C2 | C3,

6、 , CN解密運(yùn)算程序：SR1 = IV P1 = Fj (DK(SR1) C1 SRm = Sj(SRm-1) | Cm-1 ;m = 2, 3, 4, , N Pm = Fj (DK(SRm) Cm ;m = 2, 3, 4, , N P =P1 | P2 | P3, , PNk-位元輸出反饋(k-bits Output Feedback, OFB) 模式加密運(yùn)算乘程序：SR1 = IV O1 = Fj (EK(SR1) C1 = P1m = 2, 3, 4, , N Om = Fj (EK(SRm)m = 2, 3, 4, , N Cm = Om Pm ;m = 2, 3, 4, , N

7、 C = C1 | C2 | C3,解密運(yùn)算程序：SR1 = IV O1 = Fj (DK(SR1) P1 = O1m = 2, 3, 4, , N Om = Fj (DK(SRm)m = 2, 3, 4, , N Pm = Om Cm ;m = 2, 3, 4, , N P = P1 | P2 | P3, O1 SRm = Sj (SRm-1) | Om-1；CN C1 SRm = Sj (SRm-1) | Om-1PN5. Consider the following method of encrypting a message using CBC mode. To encrypt a m

8、essage, one uses the algorithm for doing a CBC decryption. To decrypt a message, one uses the algorithm for doing a CBC encryption. Would this work? How secure is this alternative method in comparison with the normal CBC mode?(10 points)考慮以下這種方法，通過 CBC模式加密一個(gè)消息。為了加密一個(gè)消息， 可以使用一個(gè)CBC解碼算法。為了解密一個(gè)消息，可以使用一個(gè)

9、CBC加密算法。這是怎么實(shí)現(xiàn)的呢？這種轉(zhuǎn)換方法與一般的CBC模式比起來有多安全呢？答：這種方式是可以的。這種方式不如通常的CBC模式安全，因?yàn)榻饷芩惴ㄊ?，每個(gè)密文組分別解密，再與上一個(gè)塊密文亦或就可以恢復(fù)明文。 解密算法的輸入是明文，這樣的安全性就非常低。6. What pseudo-random bit stream is generated by 64-bit OFB with a weak DES key?(10 points)Note: please refer to Kaufman3.3.6 for the definition ofweak key.通過使用一個(gè)弱的 DES密碼加密

10、而成的64位OFB,將產(chǎn)生什么偽隨機(jī) 的比特流呢？注：請(qǐng)參考Kaufman § 3.3.6相關(guān)章節(jié)弱密碼的定義答：OFB是用虛擬隨機(jī)數(shù)產(chǎn)生器加上IV與密鑰，產(chǎn)生次性冗字填充片段，即 Ex(IV), Ex(Ex (IV), Ex(Ex(Ex (IV),。一個(gè)弱 DES Key 是自己本身的反轉(zhuǎn)，對(duì)于任何一個(gè)塊來說b: Ex(b) = Dx(b). So Ex(Ex(b) = b.應(yīng)此OFB的填充序列是Ex(IV), IV, Ex(IV), IV,.。這樣的規(guī)律導(dǎo)致不安 全性。7. In RSA algorithm, is it possible for more than oned t

11、owork with a given e, p, and q? Kaufman 6.3(10points)在RSA算法中，當(dāng)我們知道e,p,g ,是否可能得到超過一個(gè) d?答：不能得到，要得到d必須知道p q n e。8. In RSA algorithm, what is the probability that something to be encrypted will not be in Z *n? Kaufman 7.11(10points)在RSA算法中，被加密的一些不屬于Z*n,什么情況下是可能發(fā)生的?答：Since Zn =n-1 and Zn * = n - p - q +

12、1Probability = (Zn - Zn * ) / Zn = ( P + q + 12) / ( n -1 )9. In the description of possible defense against Man-in-the-Middle Attack Kaufman§6.4.2, 3, it stated thatencrypting the Diffie- Hellman value with the other sides' spublic key shall prevent the attack. Why is this case assumption

13、that an attacker can encrypt whatever it wants with the other sides' public key? Kaufman(106.2points)在描述 defense against Man-in-the-Middle Attack這種可能性時(shí)，它說用另外種方面的普通密碼加密這個(gè)Diffie-Hellman 值可以預(yù)防這方面的攻擊。為什么一個(gè)攻擊可以用另一方面的普通密碼加密任何它想要的 呢？答：普通密碼加密這個(gè) Diffie-Hellman值，可以得知該文件的加密密碼方 式，從而找出規(guī)律。對(duì)其進(jìn)行解密。所以才有此說法。10. In modular arithmetic, why must the opera