CIS/CSE 774 Principles of Distributed Access Control Exam 1 October 3, Points Possible. Total 60

Size: px

Start display at page:

Download "CIS/CSE 774 Principles of Distributed Access Control Exam 1 October 3, Points Possible. Total 60"

Malcolm Conley
5 years ago
Views:

1 Name: CIS/CSE 774 Principles of Distributed Access Control Exam 1 October 3, 2013 Question Points Possible Points Received Total 60 Instructions: 1. This exam is a closed-book, closed-notes exam. 2. Legibility counts! Make sure I can read (and find!) your answers. If you need more room for an answer than that given, use the back side of the pages. Be sure to leave a note indicating where the answer is. 3. This test should have 8 pages (including this cover sheet). Let me know now if your copy does not have the correct number of pages.

2 1. (24 points) Consider the Kripke structure M = W, I, J, where: W = {t, u, x, y} I(p) = {t, x} I(q) = {u, x, y} I(r) = {t, u} J(Mo) = {(t, u), (u, u), (u, t), (x, x), (y, y)} J(Pat) = {(t, t), (x, y), (y, t), (y, u)} J(Sam) = {(t, t), (t, y), (u, u), (x, u), (y, x), (y, t)} (a) (3 points) What is the value of J(Sam Pat)? (b) (3 points) What is the value of J(Mo Pat)? (c) (18 points) For each formula that follows, give the set of worlds in W in which it is true. (You do not need to show your work.) i. (p q) r ii. p r iii. Mo says ( q) iv. Pat controls p

3 v. Pat Mo vi. Mo & Sam Sam Pat vii. Mo Pat says r viii. Pat says Pat Mo ix. Sam says (Pat says Pat Mo)

4 2. (12 points) Give a formal proof of the following derivable rule: P says (R controls ϕ) P Q says ϕ Q R P says ϕ

5 3. (12 points) Prove that the following proposed inference rule is sound: P controls ϕ 1 ϕ 2 P controls (ϕ 1 ϕ 2 ) That is, show that for all Kripke structures M, principals P, and formulas ϕ 1 ϕ 2 the following statement is true: If M = P controls ϕ 1 and M = ϕ 2, then M = P controls (ϕ 1 ϕ 2 ) as well.

6 4. (12 points) Show that the following proposed inference rule is not sound, and therefore should not be added to the logic: P controls (Q P ) That is, give a particular Kripke structure M and principals P, Q such that: M = P controls (Q P ). For maximal credit, be sure to provide calculations and explanations to support your answer.

7 For Reference A Language for Access Control 29 FIGURE 2.1 Semantics of core logic, for each M = W,I,J E M [[ p]] = I(p) E M [[ ϕ]] = W E M [[ϕ]] E M [[ϕ 1 ϕ 2 ]] = E M [[ϕ 1 ]] E M [[ϕ 2 ]] E M [[ϕ 1 ϕ 2 ]] = E M [[ϕ 1 ]] E M [[ϕ 2 ]] E M [[ϕ 1 ϕ 2 ]] = (W E M [[ϕ 1 ]]) E M [[ϕ 2 ]] E M [[ϕ 1 ϕ 2 ]] = E M [[ϕ 1 ϕ 2 ]] E M [[ϕ 2 ϕ 1 ]] W, if J(Q) J(P) E M [[P Q]] = /0, otherwise E M [[P says ϕ]] = {w J(P)(w) E M [[ϕ]]} E M [[P controls ϕ]] = E M [[(P says ϕ) ϕ]] Reasoning about Access Control 49 Propositional Variables: The truth of a propositional variable p is determined by FIGURE the 3.5 interpretation Some useful function derived I: rules a variable p is considered true in world w precisely when w I(p). Conjunction Thus, for all propositional variables p, ϕ 1 ϕ 2 ϕ 1 ϕ 2 E M [[ p]] = I(p). ϕ 1 ϕ 2 ϕ 1 ϕ 2 Simplification (1) Simplification (2) For example, if M 0 is the ϕ 1 Kripke structure W 0,I 0,J 0 from ϕ 2 Example 2.7, E M0 [[g]] = I 0 (g)={sw}. ϕ 1 Negation: ADisjunction formula with (1) form ϕ is truedisjunction in precisely(2) those worlds in which ϕ is not true. Because (by definition) ϕ 1 ϕ 2 E M [[ϕ]] is the set ofϕworlds 1 ϕ 2 in which ϕ is true, we define ϕ 1 ϕe 2 M [[ ϕ]] 2 = W E M [[ϕ]]. ϕ Modus Tollens Double negation ϕ Thus, returning to Example 2.7, 1 ϕ DisjunctiveE M0 ϕ 1 [[ g]] ϕ 2 = W 0 ϕ E 1 M0 [[g]] Hypothetical = {sw,sc,ns} ϕ 1 {sw} ϕ 2 = {sc,ns}. ϕ 2 ϕ 3 Syllogism ϕ 2 Syllogism ϕ 1 ϕ 3 Notice that E M0 [[ g]] is the set of worlds in which the children are not allowed to go outside. P controls ϕ P says ϕ Controls Conjunction: A conjunctive formula ϕ 1 ϕ 2 ϕis considered true in those worlds for which both ϕ 1 and ϕ 2 are true: that is, ϕ 1 ϕ 2 is true in those worlds w for which w E M [[ϕ 1 ]] and w E M [[ϕ 2 ]]. Thus, we can define E M [[ϕ 1 ϕ 2 ]] in Derived P Q P says ϕ Derived P Q Q controls ϕ terms of set intersection: Speaks For Q says ϕ Controls P controls ϕ E M [[ϕ 1 ϕ 2 ]] = E M [[ϕ 1 ]] E M [[ϕ 2 ]]. Says Simplification (1) P says (ϕ 1 ϕ 2 ) P says ϕ 1 Says Simplification (2) ϕ 2 P says (ϕ 1 ϕ 2 ) P says ϕ 2 FIGURE 3.6 A formal proof of Conjunction 1. ϕ 1 Assumption

8 40Access Control, Security, and Trust: A Logical ApproachDRAFT: DO NOTDISTRIBUTE FIGURE 3.1 Logical rules for the access-control logic Taut ϕ if ϕ is an instance of a prop-logic tautology Modus Ponens ϕ ϕ ϕ ϕ Says ϕ P says ϕ MP Says (P says (ϕ ϕ )) (P says ϕ P says ϕ ) Speaks For P Q (P says ϕ Q says ϕ) & Says (P & Q says ϕ) ((P says ϕ) (Q says ϕ)) Quoting (P Q says ϕ) (P says Q says ϕ) Idempotency of P P Transitivity of P Q Q R P R Monotonicity of P P Q Q P Q P Q Equivalence ϕ 1 ϕ 2 ψ[ϕ 1 /q] ψ[ϕ 2 /q] P controls ϕ def = (P says ϕ) ϕ Reasoning about Access Control 41 FIGURE 3.2 Common propositional-logic tautologies p p p ( p) p (q p) p (q p) (p q) p ( p p) p (q (p q)) (p q) (p q) (p q) (q p) (p q) (p q) ((p q) p) q ((p q) (q r)) (p r) (i.e., derives) various formulas on a piece of paper. Each rule states that, if all the premises of an inference rule have already been written down (derived), then the conclusion can also be written down (derived). Axioms can always be written down. We return to this notion of derivations in Section 3.2, where we introduce formal proofs. For now, however, we discuss each of the logical rules in turn.

Chapter 6 Rules of deductive inference Answers to select Getting familiar with exercises. Getting familiar with basic rules of inference.

Chapter 6 Rules of deductive inference Answers to select Getting familiar with exercises. Getting familiar with basic rules of inference. 1. 1. A 2. (A B) /.: B 3. B 1, 2 modus ponens 3. 1. (P & Q) 2.