NPTEL Artificial Intelligence: Knowledge Representation And Reasoning Week 5 Assignment Answers 2024

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NPTEL Artificial Intelligence: Knowledge Representation And Reasoning Week 5 Assignment Answers 2024

Q1. Select the formulas that are in CNF.

  • A ∧ B ∧ C
  • A ∨ B ∨ C
  • ¬A
  • (¬A ∨ B ∨ C) ∧ (¬C ∨ ¬D ∨ E)
  • ¬(A ∨ B ∨ C) ∧ (C ∨ D ∨ E)
Answer:- For Answer Click Here

Q2. Convert ∀x∀y{ ( A(x,y) ⊃ [B(x) ⊃ C(y)] ) ⊃ ( D(x) ⊃ [A(x,y) ⊃ C(y)] ) } to CNF.

  • ∀x∀y[¬A(x,y) ∧ B(x) ∧ C(y) ∧ ¬D(x)]
  • ∀x∀y{[C(y) ∨ ¬D(x)] ∧ [¬A(x,y) ∨ B(x) ∨ C(y) ∨ ¬D(x)] ∧ [¬A(x,y) ∨ ¬D(x)]}
  • ∀x∀y[¬A(x,y) ∨ B(x) ∨ C(y) ∨ ¬D(x)]
  • ∀x∀y[¬A(x,y) ∨ ¬B(x) ∨ C(y) ∨ D(x)]
Answer:- 

Q3. Convert ∀x[(∃y(P(x) ⊃ R(x,y))) ⊃ (Q(x) ∧ ∃z R(x,z))] to Skolem form.

  • ∀x∀y[(P(x) ⊃ R(x,y)) ⊃ (Q(x) ∧ R(x,sk(x)))]
  • ∀x[(P(x) ⊃ R(x,sk(x))) ⊃ (Q(x) ∧ R(x,sk(x)))]
  • ∀x∀y[(P(x) ∧ ¬R(x,y)) ∨ (Q(x) ∧ R(x,sk(x)))]
  • ∀x[(P(x) ∧ ¬R(x,sk(x))) ∨ (Q(x) ∧ R(x,sk(x)))]
Answer:- For Answer Click Here

Q4. Convert ∀x[(∃y(P(x) ⊃ R(x,y))) ⊃ (Q(x) ∧ ∃z R(x,z))] to clause form.

  • [P(?t) ∨ Q(?t)] ∧ [P(?w) ∨ R(?w,sk(?w))] ∧ [¬R(?x,?y) ∨ Q(?x)] ∧ [¬R(?u,?v) ∨ R(?u,sk(?u))]
  • [P(?t) ∨ Q(?t)] ∧ [P(?w) ∨ R(?w,sk(?w))] ∧ [¬R(?x,sk(?x)) ∨ Q(?x)] ∧ [¬R(?u,sk(?u)) ∨ R(?u,sk(?u))]
  • { {P(?t) Q(?t)}; {P(?w), R(?w,sk(?w))}; {¬R(?x,?y), Q(?x)}; {¬R(?u,?v), R(?u,sk(?u))} }
  • { {P(?t), Q(?t)}; {P(?w), R(?w,sk(?w))}; {¬R(?x,sk(?x)), Q(?x)}; {¬R(?u,sk(?u)), R(?u,sk(?u))} }
  • None of the above
Answer:- 

Q5. Consider the following pair of clauses with predicates Q, R; functions f, g; variables x, y; and constants a, b.
1. { R(f(a), g(b)) }
2. { ¬R(?x, ?y); ¬R(f(?x), g(?y)); Q(f(?x)); }
Which of the following are unifiers for the pair of clauses?

  • { x -> a; y -> b }
  • { x -> f(a); y -> b }
  • { x -> f(a); y -> g(b) }
  • None of the above because clauses with multiple unifiers cannot be unified.
Answer:- For Answer Click Here

Q6. For the pair of clauses in the previous question, select the most general unifier.
{ x -> a; y -> b }
{ x -> f(a); y -> b }
{ x -> f(a); y -> g(b) }
None of the above because clauses with multiple unifiers cannot be unified.

Answer:- 

Q7. Use the Resolution Method to derive all possible resolvents from KB1.
KB1 = { (A ∧ B) ⊃ C; Q ⊃ (P ⊃ B); D ⊃ (P ∧ Q); ¬(A ⊃ ¬D); }
Which of the following occur as a result of a resolution step?

  • ¬A ∨ ¬B ∨ C
  • B ∨ ¬P ∨ ¬Q
  • ¬B ∨ Q
  • C ∨ ¬P ∨ ¬Q
  • C ∨ ¬Q
Answer:- For Answer Click Here

Q8. Is KB1 consistent?

  • Yes
  • No
  • Cannot be determined
Answer:- 

Q9. Does KB1 entail (B ∧ C ∧ P ∧ Q)? Use the Resolution Refutation Method to prove/disprove it.

  • Yes
  • No
  • Cannot be determined
Answer:- For Answer Click Here

Q10. A KB is given in clause form with predicates A, B, P, Q, R, S; constants Foo, Bar; and variables prefixed with ‘?’.

KB2 = {
1. ¬A(?x) ∨ S(?x);
2. ¬B(?u) ∨ R(?u,?v);
3. ¬P(?y,?z) ∨ Q(?y,?z);
4. ¬S(?t) ∨ B(?t) ∨ P(?t,?w);
5. A(Foo);
6. B(Foo);
}

Which of these sentences follow from KB2. Use the Resolution Refutation Method to prove it.

  • S(Foo) ∧ S(Bar)
  • R(Foo,Bar) ∨ Q(Foo,Bar)
  • ∃t∃x∃y (R(t,x) ∨ Q(t,y))
  • ∃t∃x∃y (R(t,x) ∧ Q(t,y))
Answer:- For Answer Click Here

Q11. Can the Resolution Refutation Method be used to prove that KB2 entails the following statement?

    ∀t∃x∃y [ A(t) ⊃ (R(t,x) ∨ Q(t,y)) ]
  • Yes
  • No
Answer:- For Answer Click Here