网络安全与密码学题目
计算机程序设计员实操考核网络安全题目
计算机程序设计员实操考核网络安全题目1. 简介网络安全是计算机程序设计员需要掌握的重要技能之一。
本文档将介绍一些实操考核题目，帮助程序设计员加强对网络安全的理解和实践能力。
这些题目主要涵盖以下几个方面：1.网络安全基础知识2.网络攻击与防御3.密码学和加密算法4.数据安全和隐私保护程序设计员需要通过解答这些题目来评估自己在网络安全方面的能力，并根据答题结果来改进自己的技能。
2. 网络安全基础知识2.1 什么是网络安全？网络安全是指保护计算机网络系统不受未经授权的存取、使用、泄漏、破坏、干扰及更改等威胁和危险的技术、措施、规范和方法。
2.2 计算机网络安全的威胁类型有哪些？计算机网络安全的威胁类型包括但不限于以下几个：•病毒和恶意软件•网络钓鱼•DoS/DDoS攻击•数据泄露和信息窃取•网络入侵和黑客攻击•身份盗窃和个人隐私泄漏2.3 如何保护计算机网络安全？保护计算机网络安全需要采取一系列措施，包括但不限于以下几个方面：•使用加密技术保护数据传输和存储•安装和及时更新防火墙和杀毒软件•控制访问权限和加强身份验证•定期备份重要数据和文件•进行网络安全教育和培训3. 网络攻击与防御3.1 什么是网络攻击？网络攻击是指攻击者利用各种手段入侵计算机网络系统，获取未经授权的访问权和控制权，以实施各种违法和犯罪行为。
3.2 常见的网络攻击类型有哪些？常见的网络攻击类型包括但不限于以下几个：•网络钓鱼•病毒和恶意软件•DoS/DDoS攻击•网络入侵和黑客攻击•数据泄露和信息窃取3.3 如何进行网络防御？进行网络防御需要采取一系列措施，包括但不限于以下几个方面：•及时更新操作系统和软件漏洞补丁•使用强密码和多因素身份验证•启用防火墙和入侵检测系统•监控网络流量和日志记录•建立紧急响应机制和恢复策略4. 密码学和加密算法4.1 什么是密码学？密码学是研究信息的安全性和加密技术的科学，主要包括对称加密算法、非对称加密算法和哈希算法。
ctf新手训练题目
ctf新手训练题目CTF（Capture The Flag）是一种网络安全竞赛，旨在让参赛者通过解决一系列与网络安全相关的问题来获取旗帜（Flag）。
这些问题涵盖了密码学、漏洞利用、逆向工程、网络分析和数字取证等多个领域。
对于新手来说，CTF新手训练题目是一个很好的起点，可以帮助他们熟悉CTF竞赛的形式和解题思路。
下面是一些常见的CTF新手训练题目及其相关参考内容，希望对新手朋友们有所帮助：1. 逆向工程（Reverse Engineering）题目：逆向工程是CTF竞赛中常见的一类题目，参赛者需要破解某个程序或二进制文件，以获取其中隐藏的信息或漏洞。
参考内容：《逆向工程：初级篇》、《逆向工程：进阶篇》2. 密码学（Cryptography）题目：密码学是CTF竞赛中必不可少的一部分，参赛者需要解密密文或破解密码算法。
参考内容：《密码学入门》、《现代密码学导论》3. Web安全（Web Security）题目：Web安全是CTF竞赛中非常常见的一种题型，参赛者需要破解Web应用程序的漏洞，获得Flag。
参考内容：《Web安全技术与实战》、《Web漏洞攻防实战》4. 漏洞利用（Exploitation）题目：漏洞利用是CTF竞赛中比较高级的一种题型，参赛者需要通过利用软件或系统的漏洞来获取权限或控制目标。
参考内容：《Metasploit渗透测试与渗透工具实践》、《Linux 漏洞攻击与防范》5. 隐写术（Steganography）题目：隐写术是一种隐藏信息的技术，在CTF竞赛中出现的频率也比较高。
参赛者需要分析图片、音频或视频文件中的隐藏信息。
参考内容：《隐写术实战与分析》、《隐写术入门与实战》6. 网络分析（Network Analysis）题目：网络分析是CTF竞赛中涉及到网络协议、流量分析等内容的一种题型。
参赛者需要分析网络流量、抓包数据等来获取Flag。
参考内容：《Wireshark网络分析的艺术》、《网络分析实战指南》以上只是一些CTF新手训练题目的相关参考内容，希望能够提供一些指导和启示。
计算机安全与密码学考试题目
计算机安全与密码学考试题目
本文档提供了一些计算机安全与密码学的考试题目，用于检验学生对这一领域的理解和知识。
以下是一些典型的考试题目：
1. 简要解释密码学是什么以及其在计算机安全中的作用。
2. 详细描述对称加密和非对称加密的区别，并举例说明各自适合的应用场景。
3. 解释数字签名的概念和目的，并论述其在数据完整性和认证方面的重要性。
4. 什么是公钥基础设施（PKI）？列举PKI的组成部分，并阐述其在计算机安全中的作用。
5. 解释离散对数问题在密码学中的重要性，并说明它在DH密钥交换协议和椭圆曲线密码算法中的应用。
6. 简要描述常见的网络攻击类型，包括但不限于拒绝服务攻击、中间人攻击和SQL注入攻击，并提供防范措施。
7. 讨论隐私保护的挑战和隐私保护技术的应用，包括数据加密、匿名化和访问控制。
8. 描述常见的身份认证机制，如密码、生物特征识别和多因素
认证，并评估它们的优缺点。
以上是一些计算机安全与密码学的考试题目示例，供参考学习
之用。
可以通过深入研究每个主题和相关的实际案例来进一步加深
对这一领域的理解和应用能力。
《密码编码学与网络安全》复习题-朱铁英2006-4-16
《密码编码学与网络安全》复习题－朱铁英2006-4-16 《计算机安全与密码学》复习题1( 信息安全(计算机安全)目标是什么,答:机密性(confidentiality):防止未经授权的信息泄漏完整性(integrity):防止未经授权的信息篡改可用性(avialbility):防止未经授权的信息和资源截留抗抵赖性、不可否认性、问责性、可说明性、可审查性(accountability): 真实性(authenticity):验证用户身份2( 理解计算安全性(即one-time pad的理论安全性)使用与消息一样长且无重复的随机密钥来加密信息，即对每个明文每次采用不同的代换表不可攻破，因为任何明文和任何密文间的映射都是随机的，密钥只使用一次3( 列出并简要定义基于攻击者所知道信息的密码分析攻击类型。
(1)、唯密文分析(攻击)，密码分析者取得一个或多个用同一密钥加密的密文;(2)、已知明文分析(攻击)，除要破译的密文外，密码分析者还取得一些用同一密钥加密的密文对;(3)、选择明文分析(攻击)，密码分析者可取得他所选择的任何明文所对应的密文(不包括他要恢复的明文)，这些密文对和要破译的密文是用同一密钥加密的;(4)、选择密文分析(攻击)，密码分析者可取得他所选择的任何密文所对应的明文(要破译的密文除外)，这些密文和明文和要破译的密文是用同一解密密钥解密的，它主要应用于公钥密码体制。
4( 传统密码算法的两种基本运算是什么,代换和置换前者是将明文中的每个元素映射成另外一个元素;后者是将明文中的元素重新排列。
5( 流密码和分组密码区别是什么,各有什么优缺点,分组密码每次处理一个输入分组，对应输出一个分组;流密码是连续地处理输入元素，每次输出一个元素流密码Stream: 每次加密数据流的一位或者一个字节。
连续处理输入分组，一次输出一个元素，速度较快6( 已知密文ILPQPUN使用的是移位密码，试解密(提示:明文为有意义的英文)。
密码编码学与网络安全第七版习题8.2
密码编码学与⽹络安全第七版习题8.2这道题是密码学的作业，开始不会写，答案只有结果⽽没有过程，后来查了⼀些资料才算是搞明⽩了。
题⽬如下：(a) 下述的伪随机数发⽣器可获得的最⼤周期是多少？X n+1=(aX n)mod24(b) 这时 a 为多少？(c) 对种⼦有什么要求？解答如下：(a)⾸先引⼊这样⼀个结论：对任意的奇数a与正整数n，有：a2n≡1(mod2n+2)。
⽤归纳法证明这个结论：1. 当n=1时，存在整数b，c，使得a2n=(2b+1)2=4b(b+1)+1=23c+1≡1(mod23)2. 假设当n=k时，命题成⽴，即a2k≡1(mod2k+2)则存在整数c，使得a2k=2k+2c+1当n=k+1时，存在整数k，b，使得a2k+1=(2k+2c+1)2=22k+4c2+2·2k+2c+1=2k+3c(2k+1c+1)+1≡1(mod2k+3)即当n=k+1时，命题成⽴。
由1，2可得，该命题成⽴。
若a与24不互素，即a为偶数，令a=2k，则a4=16k4≡0(mod24)从⽽0=X n+4≡a4X n(mod24)产⽣的第四个数之后全为0，所以a与24互素。
⼜因为a2n≡1(mod2n+2)所以a24−2=a4≡1(mod24)从⽽a4X n≡X n(mod24)即X n+4=X n所以最⼤周期为4。
(b)由(a)可知，a为奇数。
经计算，a=7,9,15时，周期为2。
a=3,5,11,13时，周期为4。
(c)种⼦必须为奇数，否则周期会不⼤于2。
Processing math: 100%。
Cryptography and Network Security 密码学与网络安全
1. The three security goals are confidentiality, integrity, and availability.‰Confidentiality means protecting confidential information.‰Integrity means that changes to the information need to be done only by authorized entities. ‰Availability means that information needs to be available to authorized entities.2. In a passive attack, the attacker’s goal is just to obtain information. This means that the attack does not modify data or harm the system. Examples of passive attacks are snooping and traffic analysis.‰An active attack may change the data or harm the system. Attacks that threaten the integrity and availability are active attacks. Examples of active attacks are modification, masquerading, replaying, repudiation, and denial of service.3. We mentioned five security services: data confidentiality, data integrity, authentication, nonrepudiation, and access control.‰Data confidentiality is to protect data from disclosure attack.‰Data integrity is to protect data from modification, insertion, deletion, and replaying.‰Authentication means to identify and authenticate the party at the other end of the line.‰Nonrepudiation protects against repudiation by either the sender or the receiver of the data.‰Access control provides protection against unauthorized access to data.4. Eight security mechanisms were discussed in this chapter. encipherment, data integrity, digital signature, authentication exchange, traffic padding, routing control, notarization, and access control.‰Encipherment provides confidentiality.‰The data integrity mechanism appends a short checkvalue to the data. The checkvalue is created by a specific process from the data itself.‰A digital signature is a means by which the sender can electronically sign the data and the receiver can electronically verify the signature.‰In authentication exchange, two entities exchange some messages to prove their identity to each other.‰Traffic padding means inserting some bogus data into the data traffic to thwart the adversary’s attempt to use the traffic analysis.‰Routing control means selecting and continuously changing different avail- able routes between the sender and the receiver to prevent the opponent from eavesdropping on a particular route.‰Notarization means selecting a third trusted party to control the communication between two entities.‰Access control uses methods to prove that a user has access right to the data or resources owned by a system.5.Cryptography,a word with origin in Greek, means “secret writing.” We used the term to refer to the science and art of transforming messages to make them secure and immune to attacks. Steganography, a word with origin in Greek, means "covered writing."Steganography refers to concealing the message itself by covering it with something else.This is cryptography. The characters in the message are not hidden; they are replaced by another characters.This is steganography. The special ink hides the actual writing on the check.This is steganography. The water marks hides the actual contents of the thesis.10. A signature on a document is like a digital signature on a message. It protects the integrity of the document, it provides authentication, and it protects nonrepudiation.CHAPTER 21. The set of integers is Z. It contains all integral numbers from negative infinity topositive infinity. The set of residues modulo n is Z n. It contains integers from 0 ton −1. The set Z has non-negative (positive and zero) and negative integers; the setZ n o map a nonnegative integer from Z to Z n, weneed to divide the integer by nfrom Z to Z n, we need to repeatedly add n to the integer to move it to the range0to n−1.2. The number 1 is an integer with only one divisor, itself. A prime has only two divisors:1 and itself. For example, the prime 7 has only two divisor 7 and 1. A com- posite hasmore than two divisors. For example, the composite 42 has several divisors: 1, 2, 3, 6,7, 14, 21, and 42.3. The greatest common divisor of two positive integers, gcd (a, b), is the largestpositive integer that divides both a and b. The Euclidean algorithm can find thegreatest common divisor of two positive integers.4. The modulo operator takes an integer a from the set Z and a positive modulus n.The operator creates a nonnegative residue, which is the remainder of dividing aby n. W e mentioned three properties for the modulo operator:First: (a + b) mod n =[(a mod n) + (b mod n)] mod nSecond: (a −b) mod n = [(a mod n) −(b mod n)] mod nThird: (a ×b) mod n =[(a mod n) ×(b mod n)] mod n5. The set Z n is the set of all positive integer between 0 and n −1. The set Z n∗is the setof all integers between 0 and n −1 that are relatively prime to n. Each element in Z nZ n∗has a multiplicative inverse. The extended Euclidean algorithm is used to find the multiplicative inverses in Z n∗.6. a. 22 mod 7 =1 b. 291 mod 42 = 39c. 84 mod 320 = 84d. 400 mod 60 = 40CHAPTER 31. Symmetric-key encipherment uses a single key for both encryption anddecryption. In addition, the encryption and decryption algorithms are inverse ofeach other.2. Traditional symmetric-key ciphers can be divided into two broad categories:substitution ciphers and transposition ciphers. A substitution cipher replaces onecharacter with another character. A transposition cipher reorders the symbols.3. Substitution ciphers can be divided into two broad categories:monoalphabetic ciphers and polyalphabetic ciphers. In monoalphabetic substitution, the relation- ship between a character in the plaintext and the characters in the ciphertext is one- to-one. In polyalphabetic substitution, the relationship between a character in the plaintext and the characters in the ciphertext is one-to-many.4. Symmetric-key ciphers can also be divided into two broad categories: stream ciphers and block ciphers. In a stream cipher, encryption and decryption are done one symbol at a time. In a block cipher, symbols in a block are encrypted together.5. In a block cipher, each character in a ciphertext block depends on all characters in the corresponding plaintext block. The cipher, therefore, is a polyalphabetic.6. The additive ciphers, multiplicative ciphers, affine ciphers, and monoalphabetic substitution cipher are some examples of monoalphabetic ciphers.7. The autokey cipher, Playfair cipher, Vigenere cipher, Hill cipher, rotor cipher, and one- time pad are some examples of polyalphabetic ciphers.8. The rail fence cipher is an example of transposition cipher.9. The keyless ciphers permute the characters by using writing plaintext in one way and reading it in another way. Another method is to divide the plaintext into groups of predetermined size, called blocks, and then use a key to permute the characters in each block separately10. Brute-force attack, statistical attack, and pattern attack are examples of attacks on traditional ciphers.11. A small private club has only 100 members. Answer the following questions:a. if all members of the club need to send secret messages to each other, the number of secret keys needed = 4950.b. If everyone trusts the president of the club, and if a member needs to send a message to another member, she can follow these steps: she first sends it to the president; the president then sends the message to the other member. In this case, the number of secret keys needed = 99(填100也正确).12. If the encryption key in a transposition cipher is (3,2,6,1,5,4), the decryption key is (4,2,1,6,5,3). CHAPTER 51. The traditional symmetric-keyciphers. The2. To be resistant to exhaustive-search attack, a modern block cipher needs to be designed as a substitution cipher because in a the same number of 1s in the plaintext which makes exhaustive-search attack simpler.3. A P-box (permutation box) transposes bits. We have three types of P-boxes in modern block ciphers: straight P-boxes, expansion P-boxes, and compression P- boxes. A straight P-box is invertible; the other two are not. A straight P-box is a P-box with n inputs and m outputs where m = n. A compression P-box is a P-box with n inputs and m outputs where m < n. An expansion P-box is a P-box with n inputs and m outputs where m > n.4. An S-box is an m × n substitution unit, where m and n are not necessarily the same.The necessary condition for invertibility is that m should be equal to n .5. A product cipher is a complex cipher combining substitution, permutation , and other components discussed in this chapter.6. Diffusion hides the relationship between the ciphertext and the plaintext. Confusion hides the relationship between the ciphertext and the key.7.a. CircularLeftShift3(10011011) →11011100b. CircularRightShift3(11011100)→10011011c. The original word in Part a and the result of Part b are the same, which shows that circular left shift and circular right shift operations are inverses of each other.8. a. Swap (10011011) →10111001b. Swap (10111001) →10011011c. The original word in Part a and the result of Part b are the same, which shows that swapping is a self-invertible operation.9. Find the result of the following operations: .a. (01001101) ⊕(01001101) = (00000000)b. (01001101) ⊕(10110010) = (11111111)c. (01001101) ⊕(00000000) = (01001101)d. (01001101) ⊕(11111111) = (10110010)10. The block size in DES is 64 bits. The cipher key size is 56 bits. The round key sizeis 48 bits.11. DES uses 16 rounds.12. Double DES is vulnerable to meet-in- the-middle attack.13. Triple DES uses three stages of DES for encryption and decryption.14. The initial and final permutations are straight P-boxes that are inverses ofeach other. They have no cryptography significance in DES.15. A round in DES, as shown in this figure.L i = R i-1R i = L i-1 f(R i-1,K i)CHAPTER 81. In the electronic codebook (ECB)mode, the plaintext is divided into Nblocks.Each block is n bits. The same key is used to encrypt and decrypt each block.Advantages. This mode has two obvious advantages. First, it is simple. Second,a single bit error in transmission can create errors in several in thecorresponding block; However, the error does not have any effect on theother blocks.Disadvantages. This mode has some security problems. First, patterns at theblock level are preserved. Second, block independency creates opportunitiesfor Eve to substitutes some cipher blocks with some cipher blocks of her own.4. In the cipher block chaining (CBC) mode, each plaintext block is exclusive-oredwith the previous ciphertext block before being encrypted. A phony block called the initial vector (IV)is used to serve as C0. The same key is used to encrypt and decrypt each block.Advantages. This mode has one obvious advantage. Patterns at the blocklevel are not preserved.Disadvantages. The mode has some error-propagation problem: a single biterror in one ciphertext block C j may create errors in plaintext block P j andP j+1.5.r, where r ≤n. The idea is to use DES or AES to encrypt or decrypt the contents of a shift register,S, of size n. Data encryption is done by exclusive-oring an r-bit plaintext block with r bits of the shift register. For each block, the shift register S i is made by shifting the shift register S i−1(previous shift register) r bits to the left and filling the rightmost r bits with C i−1.Advantages. One advantage of CFB is that no padding is required because the size of the blocks, r, is normally chosen to fit the data unit to be encrypted.Disadvantages. One disadvantage of CFB is that it is less efficient than CBC or ECB, because it needs to apply the encryption function of underlying block cipher for each small block of size r.6. The output feedback (OFB) mode is very similar to CFB mode, with one differ- ence:each bit in the ciphertext is independent of the previous bit or bits. This avoids error propagation. If an error occurs in transmission, it does not affect the bits that follow.Like CFB, both the sender and the receiver use the encryption algorithm.Advantages. One advantage of OFB is that no padding is required because the size of the blocks, r, is normally chosen to fit the data unit to be encrypted.Disadvantages. One disadvantage of OFB is that it is less efficient than CBC or ECB, because it needs to apply the encryption function of underlying block cipher for each small block of size r .7. In the counter (CTR) mode, there is no feedback. The pseudorandomness in the keystream is achieved using a counter. An n-bit counter is initialized to a pre- determined value (IV) and incremented based on a predefined rule (mod 2n). To provide a better randomness, the increment value can depend on the block number to be incremented. The plaintext and ciphertext block have the same block size as the underlying cipher (e.g., DEA or AES). Plaintext blocks of size n are encrypted to create ciphertext blocks of size n .Advantages. It can be used to encrypt and decrypt random-access files .Disadvantages. It cannot be used for real-time processing. The encryption algorithm needs to wait to get a complete n-bit block of data before encrypting.8. Divide the five modes of operation into two groups: those that use the encryption and decryption functions of the underlying cipher (for example, DES) and those that use only the encryption function:9. Divide the five modes of operation into two groups: those that need padding and those that do not.10. Divide the five modes of operation into two groups: those that use the same key for the encipherment of all blocks, and those that use a key stream for encipherment of blocks.First Group: ECB and CBC Second Group: CFB, OFB, and CTRCHAPTER 101. Symmetric-key cryptography is based on sharing secrecy; asymmetric-keycryptography is based on personal secrecy.2. In asymmetric-key cryptography, each entity has a pair of public/private key. Thepublic key is universal; the private key is personal. In symmetric-keycryptography a shared secret key is used for secret communication between twoentities.3. Unlike in symmetric-key cryptography, plaintext and ciphertext are treated asintegers in asymmetric-key cryptography.4. The main idea behind asymmetric-key cryptography is the concept of the trapdoorone-way function.5. A one-way function (OWF) is a function f that satisfies the following twoproperties:(1)f is easy to compute.(2) f −1is difficult to compute.A trapdoor one-way function is a one-way function with a third property:(3) Given y and a trapdoor , x can be computed easily.6. RSA uses two exponents, e and d, where e is public and d is private. Alicecalculates C = P e mod n to create ciphertext C from plaintext P; Bob uses P =C d mod n to retrieve the plaintext sent by Alice.a. The one-way function is the C =P e mod n. Given P and e, it is easy to calculate C;given C and e, it difficult to calculate P if n is very large.b. The trapdoor in this system is the value of d, which enables Bob to use P = C dmod n.c. The public key is the tuple (e, n). The private key is d .d. The security of RSA mainly depend on the factorization of n. If n is verylarge, and the value of e and d are chosen properly, the system is secure.7. Algorithm of RSA key generation:RSA_Key_Generation{Select two large primes p and q such that p ≠qn ←p*qΦ(n) ← (p-1)*(q-1)Select e such that 1<e<Φ(n) and e is coprime to Φ(n)d ← e-1modΦ(n)Public_key ← (e,n)Private_key ← dReturn Public_key and Private_key}CHAPTER 11 Message Integrity and Message Authentication1. The difference between message integrity and message authentication is: messageintegrity guarantees that the message has not been changed; messageauthentication guarantees that the sender of the message is authentic.2. The electronic equivalent of the document and fingerprint pair is the messageand digest pair. The cryptographic hash function creates a compressedimage of the message that can be used like a fingerprint and is called messagedigest .3. The two pairs (document / fingerprint) and (message / message digest) are similar,with some differences. The document and fingerprint are physically linkedtogether. The message and message digest can be unlinked separately, and, mostimportantly, the message digest needs to be safe from change.4. To check the integrity of a message, or document, we run the cryptographic hashfunction again and compare the new message digest with the previous one. If bothare the same, we are sure that the original message has not been changed.5. A cryptographic hash function must satisfy three criteria: preimage resistance,second preimage resistance, and collision resistance. The first criteria is preimageresistance, which ensures that Eve cannot find any message whose hash is the sameas the one intercepted.The second preimage resistance ensures that a message cannoteasily be forged. In other words, given a specific message and its digest, it isimpossible to create another message with the same digest. The third criterion iscollision resistance,which ensures that Eve cannot find two messages that hash to thesame digest.6. The pigeonhole principle says that if n pigeonholes are occupied by n + 1 pigeons,then at least one pigeonhole is occupied by two pigeons. Because the digest isshorter than the message, according to the pigeonhole principle there can be collisions.In other words, there are some digests that correspond to more than one message; therelationship between the possible messages and possible digests is many-to-one.7. The following briefly states the four birthday problems:‰Problem 1: What is the minimum number, k, of students in a classroomsuch that it is likely that at least one student has a predefined birthday?‰Problem 2: What is the minimum number, k, of students in a classroomsuch that it is likely that at least one student has the same birthday as thestudent selected by the professor?‰Problem 3: What is the minimum number, k, of students in a classroom such that it is likely that at least two students have the same birthday?‰Problem 4: We have two classes, each with k students. What is the minimum value of k so that it is likely that at least one student from the firstclassroom has the same birthday as a student from the second classroom?8. The following table shows the association:9. A modification detection code is a message digest that can prove the integrity ofthe message. A message authentication code (MAC) ensures the integrity of themessage and the data origin authentication. The difference between an MDC anda MAC is that the second includes a secret between Alice and Bob. CHAPTER 13Digital SignatureReview Questions1. We mentioned four areas in which there is a difference between a conventional and a digital signature: inclusion, verification method, relationship, and duplicity.a. Inclusion: a conventional signature is included in the document; a digital signa- ture is a separate document.b. Verification method: A conventional signature is verified by comparing with the signa- ture on file. The verifier of a digital signature needs to create a new signature.c. Relationship: A document and a conventional signature have a one-to-many relation;a message and a digital signature has one-to-one relation.d. Duplicity: In conventional signature, a copy of the signed document can be distinguished from the original one on file. In digital signature, there is no such distinction unless there is a factor of time (such as a timestamp) on the document.2. A digital signature can provide three security services: message authentication,message integrity, and nonrepudiation. It does not provide confidentiality/confidential communication(斜线前后部分二选一均可算正确答案).3. A cryptosystem uses the private and public keys of the receiver; a digital signature uses the privateand public keys of the sender.4. A digital signature needs a public-key system. The signer signs with her private key; the verifier verifies with the signer’s public key.5. We discussed several security services in Chapter 1 including message confidentiality, message authentication, message integrity, and nonrepudiation.A digital signature can directly provide the last three; for message confidentiality we still need encryption/decryption.CHAPTER 14Entity AuthenticationReview Questions1.There are two differences between message authentication and entityauthentication. First, message authentication might not happen in real time;entity authentication does. Second, message authentication simplyauthenticates one message; the process needs to be repeated for each newmessage. Entity authentication authenticates the claimant for the entireduration of a session.2.Entity authentication is a technique designed to let one party prove theidentity of another party. An entity can be a person, a process, a client, or aserver. The entity whose identity needs to be proved is called the claimant;the party that tries to prove the identity of the claimant is called the verifier.3.Verification can be done with one of three kinds of witnesses: somethingknown, something possessed, or something inherent. Something known isa secret known only by the claimant that can be checked by the verifier.Something possessed is something that can prove the claimant’s identity.Something inherent is an inherent characteristic of the claimant.4.The simplest and oldest method of entity authentication is thepassword-based authentication, where the password is something that theclaimant knows. We can divide these authentication schemes into twogroups: the fixed password and the one-time password.A fixed password isa password that is used over and over again for every access. A o ne-timepassword is a password that is used only once.5.In password authentication, the claimant proves her identity bydemonstrating that she knows a secre t, the passwor d. Inchallenge-response authentication, the claimant proves that she knows asecret without sending it to the verifier.6.The challenge is a time-varying value sent by the verifier; the response isthe result of a function applied on the challenge.7.Several approaches to challenge-response authentication use symmetric-keyencryption. The secret here is the shared secret key, known by both theclaimant and the verifier. The function is the encrypting algorithm appliedon the challenge.8.A nonce is a random number used only once. A nonce must be time-varying;every time it is created, it should be different.9.In a dictionary attack, Eve is interested in finding one password, regardlessof the user ID. Eve can create a list of numbers. She then applies the hashfunction to every number until she finds a match with a hashed password.10.Guessing attacks can be prevented if people use long passwords with norelation with their names, dates of birth, account numbers, and so on.11.One problem with timestamp is the difficulty in synchronization -- thecomputer of the claimant and the verifier needs to be synchronized.12.We can use an asymmetric-key cipher for entity authentication. Here thesecret must be the private key of claimant. The claimant must show that she owns the private key related to the public key that is available to everyone.The response to the challenge is the decrypted challenge.13.When a digital signature is used for entity authentication, the claimant usesher private key for signing.。
中山大学密码学与网络安全期末复习题
中山大学密码学与网络安全期末复习题密码编码学与网络安全课程期末复习题(2013)1判断题1.四个主要的信息安全原则是:保密性,完整性,可用性,可追责性.()2.为了保证安全性,密码算法应该进行保密.()3.不可能存在信息理论安全的密码体制.()4.安全是永远是相对的,永远没有一劳永逸的安全防护措施.()5.一次一密体制即使用量子计算机也不能攻破.()6.估计维吉尼亚密文所用密钥字的长度的方法有Kasiski测试法和重合指数法.()7.Simmons囚徒问题说明了密码学的重要应用.()8.对称加密算法的基本原则是扩散(Di?usion)和混淆(Confusion).其中混淆是指将明文及密钥的影响尽可能迅速地散布到较多个输出的密文中.()9.拒绝服务攻击属于被动攻击的一种.()10.Vernam密码不属于序列密码.()11.现代分组密码都是乘法密码,分为Feistel密码和非Feistel密码两类,Feistel密码只可以运用不可逆成分.()12.流密码可以分为同步流密码和异步流密码,其中密钥流的产生并不是独立于明文流和密文流的流密码称为同步流密码.()13.DES算法中对明文的处理过程分3个阶段:首先是一个初始置换IP,用于重排明文分组的64比特数据.然后是具有相同功能的64轮变换,每轮中都有置换和代换运算.最后是一个逆初始置换从而产生64比特的密文.()14.AES算法的密钥长度是128位,分组长度为128位或192位或256位.()15.AES算法的分组长度可以是192比特.()16.AES算法不存在弱密钥和半弱密钥,能有效抵御目前已知的攻击.()期末复习题(2013)第2页(共22页)17.Di?e-Hellman算法的安全性基于离散对数计算的困难性,可以实现密钥交换.()18.常见的公钥密码算法有RSA算法,Di?e-Hellman算法和ElGamal算法.()19.ElGamal加密算法的安全性基于有限域上的离散对数难题.()20.流密码中如果第i个密钥比特与前i?1个明文有关则称为同步流密码.()21.公开密钥密码体制比对称密钥密码体制更为安全.()22.Tripe DES算法的加密过程就是用同一个密钥对待加密的数据执行3次DES算法的加密操作.()23.MD5是一个典型的Hash算法,输出的摘要值的长度可以是128位或者160位.()24.欧拉函数φ(300)=120.()25.Di?e-Hellman密钥交换协议的安全性是基于离散对数问题.()26.PGP协议缺省的压缩算法是ZIP,压缩后的数据由于冗余信息很少,更容易抵御密码分析类型的攻击.()27.我的数字证书是不能在网络上公开的,否则其他人可能假冒我的身份或伪造我的数字签名.() 28.在SSL握手协议的过程中,需要服务器发送自己的数字证书.()期末复习题(2013)第3页(共22页)2填空题1.信息安全中所面临的威胁攻击是多种多样的,一般将这些攻击分为两大类,记和被动攻击.其中被动攻击又分为和.2.主动攻击的特征是,被动攻击的特点是.3.密码学是研究通信安全保密的科学,它包含两个相对独立的分支,即学和学.4.一个保密系统一般是明文,密文,,,五部分组成的.5.密码学的发展过程中,两次质的飞跃分别是指1949年Shannon 发表的和1976年由和两人提出的思想.6.密码系统的分类有很多种,根据加密和解密所使用的密钥是否相同,密码系统可分为和.根据明文的处理方式,密码系统可分为和.7.完善保密性是指.8.Shannon证明了密码体制是绝对安全的.9.破译密码系统的方法有和.10.选择明文攻击是指.11.对称密码体制又称为密码体制,它包括密码和密码.12.古典密码是基于的密码,两类古典密码是密码和密码.13.代换是传统密码体制中最基本的处理技巧,按照一个明文字母是否总是被一个固定的字母代替进行划分,代换密码主要分为两类和.14.Hill密码可以有效抵御攻击,但不能抵御攻击.15.分组密码采用原则和原则来抵抗攻击者对该密码体制的统计分析.16.分组长度为n的分组密码可以看作是{0,1,...,2n?1}到其自身的一个置换,分组长度为n的理想的分组密码的密钥数为.17.有限域的特征一定是,有限域的元素的个数一定是其特征的.18.在今天看来,DES算法已经不再安全,其主要原因是.期末复习题(2013)第4页(共22页)19.DES算法存在个弱密钥和个半弱密钥.20.关于DES算法,密钥的长度(即有效位数)是位,又因其具有性使DES在选择明文攻击下所需的工作量减半.21.分组密码的加解密算法中最关键部分是非线性运算部分,在DES 加密算法的非线性运算部分称为,在AES加密算法的非线性运算部分称为.22.在高级加密标准AES规范中,分组长度是位,密钥的长度是位.23.AES算法支持可变的密钥长度,若密钥长度为256比特,则迭代轮数为,若密钥长度为192比特,则迭代轮数为.24.DES与AES有许多相同之处,也有一些不同之处,譬如AES密钥长度,而DES密钥长度;另外,DES是面向运算,而AES则是面向运算.25.随机序列应具有良好的统计特性,其中两个评价标准是和.26.产生伪随机数的方法有,和.27.序列密码的工作方式一般分为是和.28.消息认证码的作用是和.29.有一正整数除以3,7,11的余数分别为2,3,4,满足此条件的最小正整数是.30.公钥密码体制的思想是基于函数,公钥用于该函数的计算,私钥用于该函数的计算.31.1976年,W.Di?e和M.Hellman在一文中提出了的思想,从而开创了现代密码学的新领域.32.公钥密码体制的出现,解决了对称密码体制很难解决的一些问题,主要体现以下三个方面:问题,问题和问题.33.RSA的数论基础是定理,在现有的计算能力条件下,RSA密钥长度至少是位.34.公钥密码算法一般是建立在对一个特定的数学难题求解上,譬如RSA算法是基于困难性,ElGamal算法是基于的困难性.35.在数字签名方案中,不仅可以实现消息的不可否认性,而且还能实现消息的.期末复习题(2013)第5页(共22页)36.普通数字签名一般包括3个过程,分别是过程,过程和过程.37.1994年12月美国NIST正式颁布了数字签名标准DSS,它是在和数字签名方案的基础上设计的.38.群签名除具有一般数字签名的特点外,还有两个特征:即和.39.盲签名除具有一般数字签名的特点外,还有两个特征:即和.40.在PKI系统中CA中心的主要功能有.期末复习题(2013)第6页(共22页)3选择题1.信息安全的发展大致经历了三个发展阶段,目前是处于阶段.A.通信保密B.信息保障C.计算机安全D.网络安全2.机制保证只有发送方与接受方能访问消息内容.A.保密性B.鉴别C.完整性D.访问控制3.如果消息接收方要确定发送方身份,则要使用机制.A.保密性B.鉴别C.完整性D.访问控制4.机制允许某些用户进行特定访问.A.保密性B.鉴别C.完整性D.访问控制5.下面关于密码算法的阐述,是不正确的.A.对于一个安全的密码算法,即使是达不到理论上的不破的,也应当为实际上是不可破的.即是说,从截获的密文或某些已知明文密文对,要决定密钥或任意明文在计算机上是不可行的.B.系统的保密性不依赖于对加密算法的保密,而依赖于密钥的保密(Kerckho?s原则).C.对于使用公钥密码体制加密的密文,知道密钥的人,就一定能够解密.期末复习题(2013)第7页(共22页)D.数字签名的理论基础是公钥密码体制.6.1949年,发表题为《保密系统的通信理论》的文章,为密码系统建立了理论基础,从此密码学成了一门科学.A.Kerckho?sB.Di?e和HellmanC.ShannonD.Shamir7.一个密码系统至少由明文,密文,加密算法,解密算法和密钥五部分组成,而其安全性是由决定.A.加密算法B.解密算法C.加解密算法D.加解密算法8.计算和估计出破译密码系统的计算量下限,利用已有的最好方法破译它的所需要的代价超出了破译者的破译能力(如时间,空间,资金等资源),那么该密码系统的安全性是.A.无条件安全B.计算安全C.可证明安全D.实际安全9.根据密码分析者所掌握的分析资料的不同,密码分析一般可分为四类,其中攻击者所获信息量最大的是.A.唯密文攻击B.已知明文攻击C.选择明文攻击D.选择密文攻击10.国际标准化组织ISO所提出的信息系统安全体系结构中定义了种安全服务.A.8期末复习题(2013)第8页(共22页)B.7C.11D.511.国际标准化组织ISO所提出的信息系统安全体系结构中定义了种安全机制.A.8B.7C.11D.512.下列攻击属于被动攻击的是.A.窃听B.伪造攻击C.流量分析D.拒绝服务攻击13.下列攻击不属于主动攻击的是.A.窃听B.阻断C.篡改D.伪造14.下面关于密码算法的阐述,是不正确的.A.对于一个安全的密码算法,即使是达不到理论上的不破的,也应当为实际上是不可破的.即是说,从截获的密文或某些已知明文密文对,要决定密钥或任意明文在计算机上是不可行的.B.系统的保密性不依赖于对加密体制或算法的保密,而依赖于密钥(这就是著名的Kerckho?s原则).C.对于使用公钥密码体制加密的密文,知道密钥的人,就一定能够解密.D.数字签名的理论基础是公钥密码体制.15.下列古典密码算法是置换密码的是.期末复习题(2013)第9页(共22页)A.加法密码B.Hill密码C.多项式密码D.栅栏式密码16.字母频率分析法对算法最有效.A.置换密码B.单表代换密码C.多表代换密码D.序列密码17.算法抵抗频率分析攻击能力最强,而对已知明文攻击最弱.A.仿射密码B.维吉利亚密码C.希尔密码D.PlayFair密码18.在仿射密码中,P=C=Z26,假设某一仿射密码的加密变换记为e k(x)=7x+3,则其解密变换为.A.d k(y)=15y?19B.d k(y)=7y+3C.d k(y)=7y?3D.d k(y)=15y+1919.重合指数法对算法的破解最有效.A.置换密码B.单表代换密码C.多表代换密码D.序列密码20.维吉利亚密码是古典密码体制比较有代表性的一种密码,它属于.A.置换密码期末复习题(2013)第10页(共22页)B.单表代换密码C.多表代换密码D.序列密码21.差分分析是针对下面密码算法的分析方法.A.AESB.DESC.RC4D.MD522.DES加密算法采用位有效密钥.A.64B.56C.128D.16823.为保证安全性,在设计分组密码时应该考虑以下哪些问题.A.加密解密变换必须足够复杂,使攻击者除了用穷举法攻击以外,找不到其他简洁的数学破译方法.B.分组长度要足够大.C.密钥量要求足够大.D.加密/解密时间要足够长.24.DES采用了典型的Feistel结构,是一个迭代式的乘积密码结构,其算法的核心是.A.初始置换B.16轮的迭代变换C.逆初始置换D.轮密钥的产生25.记运行DES加密算法时使用的轮密钥为k1,k2,...,k16,则运行DES解密算法时第一轮使用的密钥是.期末复习题(2013)第11页(共22页)A.k1B.k8C.k16D.k426.AES每一轮变换的结构由如下四个不同的模块组成,其中是非线性模块.A.行移位B.列混淆C.字节代换D.轮密钥加27.AES算法中的大部分运算是按字节定义的,把一个字节看成是.A.整数域上的一个元素B.有限域GF(28)上的一个元素C.有限域GF(2)上的一个元素D.有限域GF(216)上的一个元素28.不能用来设计流密码的分组密码算法模式是.A.CFBB.OFBC.CBCD.CTR29.适合文件加密,而且有少量错误时不会造成同步失败,是软件加密的最好选择,这种分组密码的操作模式是指.A.电子密码本模式B.密码分组链接模式C.密码反馈模式D.输出反馈模式30.下列算法属于Hash算法的是.A.HMAC期末复习题(2013)第12页(共22页)B.IDEAC.RIPEMDD.RSA31.Kerberos是80年代中期,麻省理工学院为Athena项目开发的一个认证服务系统,其目标是把认证,记账和的功能扩展到网络环境.A.访问控制B.审计C.授权32.公钥密码学的思想最早由提出.A.EulerB.Di?e和HellmanC.FermatD.Rivest,Shamir和Adleman33.根据所依据的难解问题,除了以外,公钥密码体制分为以下分类.A.大整数分解问题(简称IFP)B.离散对数问题(简称DLP)C.椭圆曲线离散对数问题(简称ECDLP)D.生日悖论34.数字信封是用来解决.A.公钥分发问题B.私钥分发问题C.对称密钥分发问题D.数据完整性问题35.公钥密码主要用来进行数字签名,或者用于实现对称密码体制的密钥分配,而很少用于数据加密,主要原因是.A.公钥密码的密钥太短期末复习题(2013)第13页(共22页)B.公钥密码的效率较低C.公钥密码的安全性不好D.公钥密码抗攻击性较差36.下面不是Hash函数的等价提法.A.压缩信息函数B.哈希函数C.单向散列函数D.杂凑函数37.下面不是Hash函数具有的特性.B.可逆性C.压缩性D.抗碰撞性38.现代密码学中很多应用包含散列运算,而应用中不包含散列运算的是.A.消息完整性B.消息机密性C.消息认证码D.数字签名39.下面不是Hash函数的主要应用.A.文件校验B.数字签名C.数据加密D.认证协议40.MD5算法以位分组来处理输入文本.A.64B.128C.256期末复习题(2013)第14页(共22页)D.51241.SHA1接收任何长度的输入消息,并产生长度为比特的Hash值.A.64B.128C.160D.51242.分组加密算法(如AES)与散列函数算法(如SHA)的实现过称最大不同是.A.分组B.迭代D.可逆43.生日攻击是针对密码算法的分析方法.A.DESB.AESC.RC4D.MD544.下列算法不具有雪崩效应.A.DESB.RC4C.MD5D.RSA45.若Alice想向Bob分发一个会话密钥,采用ElGamal公钥加密算法,那么Alice应该选用的密钥是.A.Alice的公钥B.Alice的私钥C.Bob的公钥D.Bob的私钥期末复习题(2013)第15页(共22页)46.设在RSA的公钥密码体制中,公钥为(e,n)=(13,35),则私钥d=.A.11B.13C.15D.1747.在现有的计算能力条件下,对于非对称密码算法Elgamal,被认为是安全的最小密钥长度是.A.128位B.160位D.1024位48.通信中仅仅使用数字签名技术,不能保证的服务是.A.认证服务B.完整性服务C.保密性服务D.不可否认服务49.Alice收到Bob发给他的一个文件的签名,并要验证这个签名的有效性,那么签名验证算法需要Alice选用的密钥是.A.Alice的公钥B.Alice的私钥C.Bob的公钥D.Bob的私钥50.在普通数字签名中,签名者使用进行信息签名.A.签名者的公钥B.签名者的私钥C.签名者的公钥和私钥D.签名者的私钥期末复习题(2013)第16页(共22页)51.如果发送方用私钥加密消息,则可以实现.A.保密性B.保密性与鉴别C.保密性与完整性D.鉴别52.签名者无法知道所签消息的具体内容,即使后来签名者见到这个签名时,也不能确定当时签名的行为,这种签名称为.A.代理签名B.群签名D.盲签名53.签名者把他的签名权授给某个人,这个人代表原始签名者进行签名,这种签名称为.A.代理签名B.群签名C.多重签名D.盲签名54.PKI的主要理论基础是.A.对称密码算法B.公钥密码算法C.量子密码D.摘要算法55.PKI解决信息系统中的问题.A.身份信任B.权限管理C.安全审计D.数据加密期末复习题(2013)第17页(共22页)56.是PKI体系中最基本的元素,PKI系统所有的安全操作都是通过它来实现的.A.用户私钥B.用户身份C.数字证书D.数字签名57.一个典型的PKI应用系统包括实体.A.认证机构CAB.注册机构RAC.证书及CRL目录库D.用户端软件期末复习题(2013)第18页(共22页)4简答题1.简述密码分析者对密码系统的四种攻击.2.为什么二重DES并不像人们想象的那样可以提高密钥长度到112比特,而相当于57比特?简要说明原因.3.叙述中途相遇攻击(Meet-in-the-Middle Attack).4.简述序列密码算法和分组密码算法的不同.5.简述分组密码的五种操作模式及其特点.6.叙述如何应用费玛小定理(Fermat’s Little Theorem)来测试一个正整数是否为素数?7.叙述Miller-Rabin概率素性测试算法的工作原理.Miller-Rabin概率素性测试算法测试奇整数p的算法描述如下:Write p?1=2k m,where m is odd.Choose a random integer a,such that1≤a≤p?1.Compute b=a m mod p.If b=1mod p then Answer“p is a prime number”and QUIT.For i=0to k?1do–If b=?1mod p then Answer“p is a prime number”and QUIT.–Else b=b2mod pAnswer“p is not a prime number”and QUIT.Here the above Miller-Rabin algorithm is a yes-biased Monte Carlo algorithm for testing compositeness.Show that why it is?In other words,all yes-answers for the compositeness are always correct,but the no-answer for the compositeness(in other words,“p is a prime”)may be incorrect.So you have to prove that when the algorithm says“p is a composite”,then MUST be composite.8.简述链路加密和端对端加密的区别.9.公钥密码体制与对称密码体制相比有什么优点和不足?10.什么是单向函数,什么是单向陷门函数?期末复习题(2013)第19页(共22页)。
网络安全与密码学试题精选
网络安全与密码学试题精选随着互联网的快速发展和广泛应用，网络安全问题日益突出。
为了保护个人隐私和保密信息，密码学作为一门重要的学科被广泛应用于网络安全领域。
以下是一些网络安全与密码学方面的试题精选，旨在帮助读者深入了解并掌握相关知识。
题目一：对称加密与非对称加密的区别是什么？请举例说明。
解析：对称加密和非对称加密是常见的加密算法。
对称加密使用同一个密钥进行加密和解密，速度较快，但存在密钥分发问题；非对称加密使用一对密钥，一把用于加密，另一把用于解密，安全性高。
例如，对称加密算法中的DES（Data Encryption Standard）使用同一个密钥对数据进行加密和解密。
而非对称加密算法中的RSA （Rivest-Shamir-Adleman）使用一对密钥，公钥用于加密，私钥用于解密。
题目二：什么是数字签名？请简要描述数字签名的原理。
解析：数字签名是一种保证文件或信息完整性、真实性和不可抵赖性的技术。
其原理主要基于非对称加密和哈希函数。
数字签名包括生成签名和验证签名两个步骤。
生成签名的步骤如下：1. 使用哈希函数对要签名的文件进行计算，生成摘要（hash digest）。
2. 使用私钥对摘要进行加密，形成数字签名。
3. 将文件和数字签名一起发送给接收方。
验证签名的步骤如下：1. 接收方使用相同的哈希函数计算文件的摘要。
2. 使用发送方的公钥对数字签名进行解密，获取原始的摘要。
3. 比对接收到的摘要与计算得到的摘要是否一致，判断签名是否有效。
题目三：什么是DDoS攻击？请解释其原理并提出相应的防御策略。
解析：DDoS（Distributed Denial of Service）攻击是一种通过占用大量目标系统资源，使其无法正常提供服务的攻击手段。
其主要原理是利用大规模的计算机资源发起同时攻击，使目标系统超负荷运行，导致服务不可用。
针对DDoS攻击，可以采取以下防御策略：1. 流量清洗和过滤：部署专业的DDoS防护设备，对流量进行实时监测和清洗，屏蔽异常流量。
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1、请分别举例说明什么是保密性原则？完整性原则？认证原则？
不可抵赖原则？访问控制原则？可用性原则？为了实现这六个安全原则，主要采用哪些密码技术？
2、一般病毒、蠕虫、特洛伊木马三者之间最主要的差别是什么？
3、什么是密码技术？替换加密法与置换加密法有什么区别？请分
别举例说明替换加密法与置换加密法。
4、在保密通信中混淆与扩散有什么区别？请分别举两例加密算法
说明他们使用了混淆与扩散的技术。
5、请分别举例说明什么是流加密法与块（或分组）加密法？
6、中间人攻击的思想是什么？试分析采用Diffie-Hellman密钥交
换协议/算法进行公钥交换过程中可能存在的中间人攻击问题，要求用实例说明中间人攻击全过程。
7、试分析对称与非对称密钥加密体制的主要差别。
假设A是发送
方，B是接收方，他们希望进行安全的通信，请用对称与非对称密钥加密体制的结合给出一个有效的安全方案。
8、 RSA的真正关键是什么？为什么SHA比MD5更安全？
9、什么是PKI？请你用你自己的语言解释这个缩写词的含义。
试
举例分析一例使用PKI技术的案例。
10、给出最常用的三种认证类型及其实例。
11、什么是安全套接层（SSL）协议？它的主要作用是什么？
12、什么是PGP协议？它的主要作用是什么？请解释一下PGP的密
钥环概念。
13、你认为Linux比Windows 更安全吗？为什么操作系统会存在不
安全的威胁？
14、为什么数据库会存在不安全的威胁？
15、 Internet 的安全问题产生的主要原因是什么？
16、什么是防火墙？什么是入侵检测系统（IDS）？两者之间的主要
差别是什么？
17、什么是VPN？
18、什么是拒绝服务攻击？你认为如何阻止拒绝服务攻击？
19、什么是IP伪装攻击？
20、请给出一个需要安全多方计算的实例。