Artificial Intelligence: A Modern Approach
3rd Edition
ISBN: 9780136042594
Author: Stuart Russell, Peter Norvig
Publisher: Prentice Hall
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Expert Solution & Answer
Chapter 7, Problem 17E
a.
Explanation of Solution
Resolution
- The negated goal is ￢G.
- The last two clauses is resolved to produce ￢C and ￢D...
b.
Explanation of Solution
Clauses
- First, each 2-CNF clause has two places to put literals...
c.
Explanation of Solution
Propositional resolution
- Resolving two 2-CNF clauses cannot increase the clause size...
d.
Explanation of Solution
Argument
- First, the number of 3-CNF clauses is O(n3), so for nonpolynomial complexity on the basis of the number of different clauses...
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Students have asked these similar questions
Construct a truth table for (p ∨ ¬ q) ∨ (¬ p ∧ q)
Use the truth table that you constructed in part 1 to determine the truth value of (p ∨¬q) ∨ (¬ p ∧ q), given that p is true and q is false.
Determine whether the given statement is a tautology, contradiction, or contingency.
p V (~p V q)
~ (p ∧ q)
~p V ~q
use propositional logic to see if the argument is valid
(A ∧ B) ∧ (B → A’) → (C ∧ B’)
A ∧ B Hypotheses
B → A’ Hypotheses
Chart to go off of attached below
In Boolean logic, a sentence is in 3-Conjunctive Normal Form (abbreviated 3-CNF) if it of the form
C¡ ^ C2 ^ C3 ^ .…. A Cm, where S is the conjunction of m clauses, with each clause being a disjunction of
S
three literals.
For example, the following sentence is in 3-CNF, with m=4 clauses.
(A v B v č) ^ (A v c v D)^ (B v
vc v D)A (a v E v D)
S =
A V B V C
A V C
D) A
B V
В
For example, the first clause is equivalent to this in English: "either A or B or Not C". As long as one of these three
literals is TRUE then this clause evaluates to TRUE. If none of these literals is TRUE (i.e., A=FALSE, B=FALSE,
C=TRUE), then this clause evaluates to FALSE.
We say that S is satisfiable if there exists an assignment of TRUE/FALSE values to each variable (A, B, C, D) so that the
entire sentence S evaluates to TRUE. What this means is that each of the m clauses must evaluate to TRUE.
The above sentence indeed is satisfiable; one possible assignment is A=TRUE, B=FALSE, C=FALSE, D=TRUE.
The first two…
Chapter 7 Solutions
Artificial Intelligence: A Modern Approach
Ch. 7 - Suppose the agent has progressed to the point...Ch. 7 - (Adapted from Barwise and Etchemendy (1993).)...Ch. 7 - Prob. 3ECh. 7 - Which of the following are correct? a. False |=...Ch. 7 - Prob. 5ECh. 7 - Prob. 6ECh. 7 - Prob. 7ECh. 7 - We have defined four binary logical connectives....Ch. 7 - Prob. 9ECh. 7 - Prob. 10E
Ch. 7 - Prob. 11ECh. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - Prob. 15ECh. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - A sentence is in disjunctive normal form (DNF) if...Ch. 7 - Prob. 20ECh. 7 - Prob. 21ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27E
Knowledge Booster
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- In Boolean logic, a sentence is in 3-Conjunctive Normal Form (abbreviated 3-CNF) if it of the form C1 ^ C2 ^ C3 ^ .…. A Cm, where S is the conjunction of m clauses, with each clause being a disjunction of S three literals. We say that S is satisfiable if there exists an assignment of TRUE/FALSE values to each variable so that the entire sentence S evaluates to TRUE. For example, in the last class, we showed that the following 3-CNF sentence, with m=8 clauses, is not satisfiable. D) ^ (B v č v D) ^ (c vĀ v D) ^ (A (A v (Bv č v D) A (c v Ā v D) a (A v B V C) ^ (a v B v C) A V v B v D) S = AVВV A V B Let 3-SAT be the decision problem that inputs a 3-CNF sentence S, and outputs YES if S is satisfiable, and outputs NO if S is not satisfiable. Consider these four statements. A. 3-SAT is in P, but is almost surely not in NP B. 3-SAT is in NP, but is almost surely not in P C. 3-SAT is in P and is also in NP D. 3-SAT is neither in P nor in NP Determine which of these four statements is correct.…arrow_forwardConstruct a truth table for the given statement. Identify whether the statement is a tautology. (-q-p) → (q^p) P T T F F q (-q-p) → (q^ p) T F T F ➤ ➤ *** Is the statement a tautology? OA. The statement is not a tautology, since it is false for all combinations of truth values of the components. OB. The statement is not a tautology, since there is at least one combination of truth values for its components where the statement is false. OC. The statement is a tautology, since there is at least one combination of truth values for its components where the statement is true. OD. The statement is a tautology, since it is true for all combinations of truth values of the components.arrow_forwardPart 1: Proof-Theoretic Concepts Show that each of the following pairs of sentences are provably equivalent in SL 1. P → R, ¬R → ¬Parrow_forward
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