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# SHW3 - IS2150/TEL2810 Introduction to Security Homework 3...

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IS2150/TEL2810 Introduction to Security Homework 3 Sample Solution Question 1: Section 2.6 Exercises 1 and 2

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Question 1: Section 3.5 Exercise 1
Question 2: Exercise on Lattice S = {11, 12, 13, 21, 22, 23, 31, 32, 33} The relation lessequivlnt over set S is partial order, as it is reflexive (e.g., 11 lessequivlnt 11), anti-symmetric (e.g., 11 lessequivlnt 12 but 12 notlessequivlnt 11), and transitive (e.g., 11 lessequivlnt 12, 12 lessequivlnt 32, and 11 lessequivlnt 32). However, it is not a total order because not every pair of elements are comparable (e.g., 13 notlessequivlnt 32 and 32 notlessequivlnt 13, i.e., the relation is not applying to 13 and 32 either way). 11 12 13 21 22 32 33 31 23

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It’s also a lattice. Because in addition to being partial order (reflexive, anti-symmetric, and transitive), every two elements have a greatest lower bound and least upper bound. Question 3: Section 4.8 Exercise 3, 4, 5, and 6 3. (a) Assume that the system has no integrity controls. This is true. If a system lacks integrity, then data can be changed without restraint. So, anyone can change another user’s authentication information, allowing them access to that user’s account—and allowing them to see any data for that user or, by generalizing this in the obvious way, any user on the system. If some integrity controls work, then the ability of the system to provide confidentiality depends on the effectiveness of the integrity controls and their use to protect critical information. b. Assume that the system has no confidentiality controls. If there is no confidentiality, then all authentication information will be available. Unless authentication mechanisms do not use secret information (for example, biometrics or positions), any user can authenticate as another user. Hence there is no integrity. Now suppose authentication information does not rely on confidentiality. Can the data in a file be kept confidential? To do so, either the user must be prevented from reading the file (for which there are no controls) or from reading the data in the file (for example, by cryptography). In the latter case, if the data is encrypted on the system, the key must be available, and as there is no confidentiality, the key can be read. If the data is not encrypted on the system, then the data cannot be used on the system but will remain
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