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# Gremlins and Goblins

Gremlins and Goblins. Gremlins always tell the truth. Goblins sometimes lie. Unfortunately, their appearance is identical.

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## Gremlins and Goblins

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1. 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.

2. 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.

3. 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.

4. 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))

5. 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)

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

7. 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]

8. 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

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

10. The Answer • The exit is red!

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