answers to russell and norvig question 7.3

Intelligent Systems

Group 3: Question 7.3

7.3a) Write a recursive algorithm PL-TRUE?(s,m) that returns true if and only if the sentence

s is true in the model m (where m assigns a truth value for every symbol in s). The algorithm

should run in time linear in the size of the sentence. Assumptions: • The sentence s is made unambiguous by the use of parentheses.• For brevity, we only include the symbols ∧, ∨, and ¬• We use m[a] to denote the true/false value of a in model mfunction PL-TRUE?(s, m) returns true or false:1) Let LHS = ‘’, RHS=‘’, logic_operator = ‘’2) Scan s from left to right, looking for a ¬, left parenthesis, or symbol. 3) If symbol found, LHS = symbol, and goto line 114) If ¬ found, logic_operator = ¬. Goto line 2.5) If left parenthesis found:6) Let n = 17) While n > 0:8) LHS += next character9) if next character == right parenthesis: n = n -110) if next character == left parenthesis: n = n +111) If logic_operator == ‘’, logic_operator = next character.12) Repeat lines 3 - 11, replacing LHS with RHS13) If LHS or RHS is not a symbol, LHS = PL_TRUE?(LHS, m), RHS = PL_TRUE?(RHS, m) as needed14) If logic_operator == ¬:15) return true if LHS is false, false otherwise.16) else If logic_operator == ∧:17) return true if LHS and RHS are both true, false otherwise18) else If logic_operator == ∨:19) return true if either LHS or RHS are equal to true, false otherwise

7.3b) Give three examples of sentences that can be determined to be true or false in

a partial model that does not specify a truth value for some of the symbols.

• true• false• A ∨ true

7.3c) Show that the truth value (if any) of a sentence in a partial model cannot be

determined efficiently in general.

• Let k be the number of symbols in s that do not appear in the partial model m.

• Some sentences are always true or always false, no matter the truth-table assignment of the symbols in the sentence.

• The only way to be sure is to evaluate the 2k rows of the truth table one-by-one, waiting to see one true and one false evaluation, to determine if the sentence has indeterminate truth value.

• However, due to the existence of always true and always false sentences, it is in the worst case a O(2k) operation determining the truth value of a sentence.

7.3c) Modify your PL-TRUE? algorithm so that it can sometimes judge truth from

partial models, while retaining its recursive structure and linear run time. Give three

examples of sentences whose truth in a partial model is not detected by your algorithm. • The modification occurs when evaluating ∧, ∨

logical operations• In line 13,(If LHS or RHS is not a symbol, LHS = PL_TRUE?(LHS, m), RHS = PL_TRUE?(RHS, m) as needed)the LHS and RHS come back as variables, or sentences containing symbols not included in the partial model. • When evaluating ∨, if either the LHS or RHS

are true values, PL-TRUE? can return true• When evaluating ∧, if either of LHS or RHS

values are false, PL-TRUE? can return false• Example of sentences modified PL-TRUE?

cannot determine as true or false, if Q is not defined in the model

Q ∨ ¬QQ ∧ ¬Q¬Q

7.3c) Investigate whether the modified algorithm makes TT-ENTAILS? more efficient.

Yes, because in many cases involving the ∧ and ∨ operators, the value of the truth assignment can be returned without needing to evaluate both sides of the operator. This may further mean that there is no need to recurse.