信息安全期末考试试卷抢先版

蝎子作品 1、 What are basic components of computer security ? try to give the connotation of each item in your list. Ans: Confidentiality: Keeping data and resources hidden. Integrity : refers to the trustworthiness of data or resources, and it is usually phrased in terms of preventing improper or unauthorized change. Integrity includes data integrity and origin integrity. Availability: refers to the ability to use the information or resource desired，enabling access to data and resources.

2、What are security policy and security mechanism ? Ans: A security policy is a statement of what is, and what is not, allowed. A security mechanism is a method, tool, or procedure for enforcing a security policy.

3、What are differences between MAC and DAC ? Ppt上是What are MAC and DAC Ans:MAC: identity is irrelevant. system mechanism controls access to object, and individual cannot alter that access DAC: is based on user identity. individual user sets access control mechanism to allow or deny access to an object

4、Consider a computer system with three users: Alice, Bob and Cyndy. Alice owns the file alicerc , and Bob and Cyndy can read it .Cyndy can read and write Bob’s file bobrc , but Alice can only read it. Only Cyndy can read and write her file cyndyrc. Assume that the owner of each of these files can execute it.  create the corresponding access control matrix.  Cyndy gives Alice permission to read cyndyrc. And Alice removes Bob’s ability to read alicerc. Show the new access control matrix.  (必考) Ans: （Read、write、Owns、execute） 蝎子作品 5、Cæsar cipher & Vigènere Cipher & DES （必考） Ans： the Caesar cipher is a classical cipher. Sender, receiver share common key Keys may be the same, or trivial to derive from one another Two basic types ：Transposition ciphers and Substitution ciphers 。problem：Key is too short （Transposition ciphers ： Plaintext is HELLO WORLD Rearrange as HLOOL ELWRD Ciphertext is HLOOL ELWRD Substitution ciphers：Plaintext is HELLO WORLD Change each letter to the third letter following it (X goes to A, Y to B, Z to C) Key is 3, usually written as letter ‘D’ Ciphertext is KHOOR ZRUOG） Vigènere Cipher Like Cæsar cipher, but use a phrase Example Message THE BOY HAS THE BALL Key VIG Encipher using Cæsar cipher for each letter: key VIGVIGVIGVIGVIGV plain THEBOYHASTHEBALL cipher OPKWWECIYOPKWIRG 6、What are definitions of object and subject ?（很有可能会考） Ans: The set of all protected entities (that is, entities that are relevant to the protection state of the system) is called the set of objects O. The set of subjects S is the set of active objects, such as processes and users. Subject can be object，And not vice versa 蝎子作品 7、Specify the algorithm of public key digital signatures（数字签名）. Ans: Sender uses hash function to compress the plaintext to generate the hash value, and then sender uses the private key to encrypt the hash value, the hash value after encryption and plaintext passed to the receiver, then the receiver uses the public key of the sender to decrypt , and the receiver uses hash function to compress the plaintext, and generate another hash value, finally compare two hash values, if they equal, so it is the real signature, otherwise not. 8、List the basic requirements of cryptographic checksum function.(重点) Ans: Cryptographic checksum h: AB: For any x  A, h(x) is easy to compute For any y  B, it is computationally infeasible to find x  A such that h(x) = y It is computationally infeasible to find two inputs x, x  A such that x ≠ x and h(x) = h(x) Alternate form (stronger): Given any x  A, it is computationally infeasible to find a different x  A such that h(x) = h(x). 9、What are differences between the classical key cryptography and public key cryptography ?（必考） Ans: the classical key cryptography has one key，Sender, receiver share common key，Keys may be the same, or trivial to derive from one another. public key cryptography has Two keys，Private key known only to individual，Public key available to anyone. 10、For RSA, take prime numbers p=7,q=13, try to calculate the public key and private key . (注意点：答案可能不唯一) Ans: n=pq=91, (n) = (p–1)(q–1)=72, e*d mod (n)=1, e=5,d=29. Public key (e, n)=(5,91) private key: d=29 11、Classes of Threats Ans: • Disclosure – Snooping • Deception – Modification, spoofing, repudiation of origin, denial of receipt • Disruption – Modification • Usurpation – Modification, spoofing, delay, denial of service 12、Goals of Security Ans: • Prevention – Prevent attackers from violating security policy • Detection