Gremlins and Goblins Gremlins always tell the truth. Goblins sometimes lie. Unfortunately, their appearance is identical. You are in a dungeon with two.

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Presentation transcript:

Gremlins and Goblins Gremlins always tell the truth. Goblins sometimes lie. Unfortunately, their appearance is identical. You are in a dungeon with two gremlins and a goblin (but you don’t know which is which). There is a red door and a blue door. One door leads to the exit. The other door leads to a pit of deadly venomous snakes.

The Query Your objective is to find the exit. Your companions give you some advice: –Gorbnitz (G1): The red door is the way out. I am telling the truth! –Grovnack (G2): The blue door is the way out. Gorbnitz is lying!! –Glidnoph (G3): I don’t know which door is the way out, but Gorbnitz is telling the truth.

Solving the Problem Express the problem in first-order logic. Express the negated goal in first-order logic and add it to the KB. Translate all of the knowledge into standard form (clausal normal form). Apply resolution to derive a contradiction. The answer is in the bindings.

Describing the Problem Goblins /gremlins mutually exclusive… –  x Gob(x)   Grem(x) –  x Grem(x)   Gob(x) One goblin… –Gob(G1)  Gob(G2)  Gob(G3) Two gremlins… –(Grem(G1)  Grem(G2))  (Grem(G1)  Grem(G3))  (Grem(G1)  Grem(G3))

The Problem II Exactly one door is the exit –Exit(Red)  Exit(Blue) –  (Exit(Red)  Exit(Blue) The creatures’ statements: –Grem(G1)  Exit(Red)  Grem(G1) –Grem(G2)  Exit(Blue)  Grem(G2) –Grem(G3)  Grem(G1)

Describing the Negated Goal Goal:  e Exit(e) Negated goal:  e  Exit(e)

Clausal Normal Form  Gob(x)   Grem(x)  Grem(y)   Gob(y) [redundant] Gob(G1)  Gob(G2)  Gob(G3) Grem(G1)  Grem(G2) Grem(G2)  Grem(G3) Grem(G1)  Grem(G3)  Grem(G1)  Exit(Red)  Grem(G1)  Grem(G1) [tautology]  Grem(G2)  Exit(Blue)  Grem(G2)  Grem(G2) [tautology]  Grem(G3)  Grem(G1)  Exit(e) [negated goal]

Think about the proof! Both G1 and G3 say that G1 is telling the truth, so this must be true (since they can’t both be lying) If G1 is telling the truth, then the red door is the exit! Now all we have to do is construct the resolution proof tree that demonstrates this line of reasoning

Resolution Proof Tree  Grem(G3)  Grem(G1)Grem(G3)  Grem(G1) Grem(G1)  Grem(G1)  Exit(Red) Exit(Red)  Exit(e) FALSE {e/Red}

The Answer The exit is red!