网络安全技术英文习习题集_网络安全技术

1、网络安全技术英文习题集Chapter 1 IntroductionANSWERS NSWERS TO QUESTIONS What is the OSI security architectureThe OSI Security Architecture is a framework that provides a systematic way of defining the requirements for security and characterizing the approaches to satisfying those requirements. The document def

2、ines security attacks, mechanisms, and services, and the relationships among these categories. What is the difference between passive and active security threatsPassive attacks have to do with eavesdropping on, or monitoring, transmissions. Electronic mail, file transfers, and client/server exchange

3、s 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. Lists and briefly define categories of passive and active security attacksPassive attacks: release of message contents and

4、traffic analysis. Active attacks: masquerade, replay, modification of messages, and denial of service. Lists and briefly define categories of security serviceAuthentication: The assurance that the communicating entity is the one that it claims to be. Access control: The prevention of unauthorized us

5、e of a resource ., this service controls who can 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

6、exactly as sent by an authorized entity ., 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

7、 system or a system resource being accessible and usable upon demand by an authorized system entity, according to performance specifications for the system ., a system is available if it provides services according to the system design whenever users request them).Chapter2 Symmetric Encryptionand Me

8、ssage ConfidentialityANSWERS NSWERS TO QUESTIONS What are the essential ingredients of a symmetric cipherPlaintext, encryption algorithm, secret key, ciphertext, decryption algorithm. What are the two basic functions used in encryption algorithmsPermutation and substitution. How many keys are requir

9、ed for two people to communicate via a symmetric cipherOne secret key. What is the difference between a block cipher and a stream cipherA 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

10、and used to produce a ciphertext block of equal length. What are the two general approaches to attacking a cipherCryptanalysis and brute force. Why do some block cipher modes of operation only use encryption while others use both encryption and decryptionIn some modes, the plaintext does not pass th

11、rough the encryption function, but is XORed with the output of the encryption function. The math works out that for decryption in these cases, the encryption function must also be used. What is triple encryptionWith triple encryption, a plaintext block is encrypted by passing it through an encryptio

12、n algorithm; the result is then passed through the same encryption algorithm again; the result of the second encryption is passed through the same encryption algorithm a third time. Typically, the second stage uses the decryption algorithm rather than the encryption algorithm. Why is the middle port

13、ion of 3DES a decryption rather than an encryptionThere is no cryptographic significance to the use of decryption for the secondstage. Its only advantage is that it allows users of 3DES to decrypt data encrypted by users of the older single DES by repeating the key. What is the difference between li

14、nk and end-to-end encryptionWith link encryption, each vulnerable communications link is equipped on both ends with an encryption device. With end-to-end encryption, the encryption process is carried out at the two end systems. The source host or terminal encrypts the data; the data in encrypted for

15、m are then transmitted unaltered across the network to the destination terminal or host. List ways in which secret keys can be distributed to two communicating parties.For two parties A and B, key distribution can be achieved in a number of ways, as follows:(1)A can select a key and physically deliv

16、er it to B.(2)A third party can select the key and physically deliver it to A and B.(3)If A and B have previously and recently used a key, one party can transmit the new key to the other, encrypted using the old key.(4)If A and B each has an encrypted connection to a third party C, C can deliver a k

17、ey on the encrypted links to A and B. What is the difference between a session key and a master keyA session key is a temporary encryption key used between two principals. A master key is a long-lasting key that is used between a key distribution center and a principal for the purpose of encoding th

18、e transmission of session keys. Typically, the master keys are distributed by noncryptographic means. What is a key distribution centerA key distribution center is a system that is authorized to transmit temporary session keys to principals. Each session key is transmitted in encrypted form, using a

19、 master key that the key distribution center shares with the target principal.ANSWERS NSWERS TO PROBLEMS What RC4 key value will leave S unchanged during initialization That is, after the initial permutation of S, the entries of S will be equal to the values from 0 through 255 in ascending order.Use

20、 a key of length 255 bytes. The first two bytes are zero; that is K0 = K1 = 0. Thereafter, we have: K2 = 255; K3 = 254; K255= 2. If a bit error occurs in the transmission of a ciphertext character in 8-bit CFB mode, how far does the error propagate Nine plaintext characters are affected. The plainte

21、xt character corresponding to the ciphertext character is obviously altered. In addition, the altered ciphertext character enters the shift register and is not removed until the next eight characters are processed. Key distribution schemes using an access control center and/or a key distribution cen

22、ter have central points vulnerable to attack. Discuss the security implications of such centralization.The central points should be highly fault-tolerant, should be physically secured, and should use trusted hardware/software.Chapter 3 Public-Key Cryptography and Message AuthenticationANSWERS NSWERS

23、 TO QUESTIONS List three approaches to message authentication.Message encryption, message authentication code, hash function. What is message authentication codeAn authenticator that is a cryptographic function of both the data to be authenticated and a secret key. Briefly describe the three schemes

24、 illustrated in .(a) A hash code is computed from the source message, encrypted using symmetric encryption and a secret key, and appended to the message. At the receiver, the same hash code is computed. The incoming code is decrypted using the same key and compared with the computed hash code. (b) T

25、his is the same procedure as in (a) except that public-key encryption is used; the sender encrypts the hash code with the sender's private key, and the receiver decrypts the hash code with the sender's public key. (c) A secret value is appended to a message and then a hash code is calculated

26、 using the message plus secret value as input. Then the message (without the secret value) and the hash code are transmitted. The receiver appends the same secret value to the message and computes the hash value over the message plus secret value. This is then compared to the received hash code. Wha

27、t properties must a hash function have to be useful for message authentication(1)H can be applied to a block of data of any size.(2)H produces a fixed-length output.(3)H(x) is relatively easy to compute for any given x, making both hardware and software implementations practical.(4)For any given val

28、ue h, it is computationally infeasible to find x such that H(x) = h. This is sometimes referred to in the literature as the one-way property.(5)For any given block x, it is computationally infeasible to find y x with H(y) =H(x).(6)It is computationally infeasible to find any pair (x, y) such that H(

29、x) = H(y).3.5 In the context of a hash function, what is a compression functionThe compression function is the fundamental module, or basic building block, of a hash function. The hash function consists of iterated application of the compression function. What are the principal ingredients of a publ

30、ic-key cryptosystemPlaintext: This is the readable message or data that is fed into the algorithm as input. Encryption algorithm: The encryption algorithm performs various transformations on the plaintext. Public and private keys: This is a pair of keys that have been selected so that if one is used

31、 for encryption, the other is used for decryption. The exact transformations performed by the encryption algorithm depend on the public or private key that is provided as input. Ciphertext: This is the scrambled message produced as output. It depends on the plaintext and the key. For a given message

32、, two different keys will produce two different ciphertexts. Decryption algorithm: This algorithm accepts the ciphertext and the matching key and produces the original plaintext. List and briefly define three uses of a public-key cryptosystem.Encryption/decryption: The sender encrypts a message with

33、 the recipient's public key. Digital signature: The sender "signs" a message with its private key. Signing is achieved by a cryptographic algorithm applied to the message or to a small block of data that is a function of the message. Key exchange: Two sides cooperate to exchange a sess

34、ion key. Several different approaches are possible, involving the private key(s) of one or both parties. What is the difference between a private key and a secret keyThe key used in conventional encryption is typically referred to as a secret key. The two keys used for public-key encryption are refe

35、rred to as the public key and the private key. What is digital signatureA digital signature is an authentication mechanism that enables the creator of a message to attach a code that acts as a signature. The signature is formed by taking the hash of the message and encrypting the message with the cr

36、eator's private key. The signature guarantees the source and integrity of the message. What is a public-key certificateA pubic-key certificate consists of a public key plus a User ID of the key owner, with the whole block signed by a trusted third party. Typically, the third party is a certifica

37、te authority (CA) that is trusted by the user community, such as a government agency or a financial institution. How can public-key encryption be used to distribute a secret keySeveral different approaches are possible, involving the private key(s) of one or both parties. One approach is Diffie-Hell

38、man key exchange. Another approach is for the sender to encrypt a secret key with the recipient's public key.ANSWERS NSWERS TO PROBLEMS Consider a 32-bit hash function defined as the concatenation of two 16-bit functions: XOR and RXOR, defined in Section as “two simple hash function.”a. Will thi

39、s checksum detect all errors caused by an odd number of error bits Explain.b. Will this checksum detect all errors caused by an even number of error bits If not, characterize the error patterns that will cause the checksum to fail.c. Comments on the effectiveness of this function for use a hash func

40、tions for authentication.a. Yes. The XOR function is simply a vertical parity check. If there is an odd number of errors, then there must be at least one column that contains an odd number of errors, and the parity bit for that column will detect the error. Note that the RXOR function also catches a

41、ll errors caused by an odd number of error bits. Each RXOR bit is a function of a unique "spiral" of bits in the block of data. If there is an odd number of errors, then there must be at least one spiral that contains an odd number of errors, and the parity bit for that spiral will detect

42、the error.b. No. The checksum will fail to detect an even number of errors when both the XOR and RXOR functions fail. In order for both to fail, the pattern of error bits must be at intersection points between parity spirals and parity columns such that there is an even number of error bits in each

43、parity column and an even number of error bits in each spiral.c. It is too simple to be used as a secure hash function; finding multiple messages with the same hash function would be too easy. Suppose H (m) is a collision resistant hash function that maps a message of arbitrary bit length into an n-

44、bit hash value. Is it true that, for all messages x, x with xx,we have H(x)H(x)Explain your answer.The statement is false. Such a function cannot be one-to-one because the number of inputs to the function is of arbitrary, but the number of unique outputs is 2n. Thus, there are multiple inputs that m

45、ap into the same output. Perform encryption and decryption using the RSA algorithm, as in , for the following:a. p=3;q=11;e=7;M=5b. p=5;q=11;e=3;M=9c. p=7;q=11;e=17;M=8d. p=11;q=13;e=11;M=7e. p=17;q=31;e=7;M=: Decryption is not as hard as you think; use some finesse.a. n = 33; ö(n) = 20; d = 3;

46、 C = 26.b. n = 55; ö(n) = 40; d = 27; C = 14.c. n = 77; ö(n) = 60; d = 53; C = 57.d. n = 143; ö(n) = 120; d = 11; C = 106.e. n = 527; ö(n) = 480; d = 343; C = 128. For decryption, we have128343 mod 527 = 128256 × 12864 × 12816 × 1284 × 1282 × 1281 mod 527

47、= 35 × 256 × 35 × 101 × 47 × 128 = 2 mod 527= 2 mod 2573.4 In a public-key system using RSA, you intercept the cipher text C=10 sent to a user whose public key is e=5, n= is the plaintext MM = 53.5 In an RSA system, the public key of a given user is e=31, n= is the private k

48、ey of this userd = 3031 Suppose we have a set of blocks encoded with the RSA algorithm and we dont have the private key, Assume n=pq, e is the public key. Suppose also someone tells us they know one of the plaintext blocks has a common factor with n. Does this help us in any wayYes. If a plaintext b

49、lock has a common factor with n modulo n then the encoded block will also have a common factor with n modulo n. Because we encode blocks that are smaller than pq, the factor must be p or q and the plaintext block must be a multiple of p or q. We can test each block for primality. If prime, it is p o

50、r q. In this case we divide into n to find the other factor. If not prime, we factor it and try the factors as divisors of n. Consider a Diffie-Hellman scheme with a common prime q=11 and a primitive root a=2.a. If user A has public key YA=9, what is As private key XAb. If user B has public key YB=3

51、, what is the shared secret key Ka. XA = 6b. K = 3Chapter 4 Authentication ApplicationsANSWERS NSWERS TO QUESTIONS What problem was Kerberos designed to addressThe problem that Kerberos addresses is this: Assume an open distributed environment in which users at workstations wish to access services o

52、n servers distributed throughout the network. We would like for servers to be able to restrict access to authorized users and to be able to authenticate requests for service. In this environment, a workstation cannot be trusted to identify its users correctly to network services. What are three thre

53、ats associated with user authentication over a network or InternetA user may gain access to a particular workstation and pretend to be another user operating from that workstation. 2. A user may alter the network address of a workstation so that the requests sent from the altered workstation appear

54、to come from the impersonated workstation. 3. A user may eavesdrop on exchanges and use a replay attack to gain entrance to a server or to disrupt operations. List three approaches to secure user authentication in a distributed environment.Rely on each individual client workstation to assure the ide

55、ntity of its user or users and rely on each server to enforce a security policy based on user identification (ID). 2. Require that client systems authenticate themselves to servers, but trust the client system concerning the identity of its user. 3. Require the user to prove identity for each servic

56、e invoked. Also require that servers prove their identity to clients. What four requirements are defined for KerberosSecure: A network eavesdropper should not be able to obtain the necessary information to impersonate a user. More generally, Kerberos should be strong enough that a potential opponent

57、 does not find it to be the weak link. Reliable: For all services that rely on Kerberos for access control, lack of availability of the Kerberos service means lack of availability of the supported services. Hence, Kerberos should be highly reliable and should employ a distributed server architecture

58、, with one system able to back up another. Transparent: Ideally, the user should not be aware that authentication is taking place, beyond the requirement to enter a password. Scalable: The system should be capable of supporting large numbers of clients and servers. This suggests a modular, distribut

59、ed architecture. What entities constitute a full-service Kerberos environmentA full-service Kerberos environment consists of a Kerberos server, a number of clients, and a number of application servers.4.6 In the context of Kerberos, what is a realmA realm is an environment in which: 1. The Kerberos server must