A Baron, a Count, a Duke and an Earl met at a jousting tournament. In the first round, two met in the first joust, and the other two met in the second joust; the two winners from the first round met at the second round for the final joust. After the jousting, they declared:

Baron: I beat the Earl.

Count: I faced both the Baron and the Duke.

Duke: I didn't make it past the first round.

Earl: At the first round, I lost to the Duke.

I knew how many were knights, and how many were liars (though not who was what) but that wasn't enough to know what jousts there had been.

However, I happened to know that a certain joust had taken place (though I didn't know who won and if it had been in the first or the second round) and that allowed me to know every result.

Can you deduce this?

In this unusual situation, the speaker tells us that knowing how many knights and liars is not enough to know which jousts took place, but knowing that a specific joust took place without knowing who won is enough to deduce the full results of the tourney. The only way knowing that a specific joust took place can give us all of the results is if that knowledge implies that the Baron (B) faced E in the SECOND round, AND that B was a knight. Otherwise, there'd be no way to determine the outcome of the final round since no other statement, true or false, can require a specific outcome in round 2.

So the indecision of the speaker requires B facing E in round 2 and winning. Since D therefore lost in round one, D told the truth and is also a Knight. And since E won in round one, E is a liar. Finally, since C also lost in round one, C is a liar for a grand total of two liars and two truth tellers. So we know the speaker knew that there were exactly two liars.

There are two scenarios for round one consistent with what we know so far: (1) B beats D, E beats C; (2) B beats C, E beats D. Hold that thought, and now work in the other direction.

What are the possible scenarios where we *only* use the fact that there are exactly two liars? Well, if E is a knight, the liars are B and D (they both contradict E's true statement) and we'd have round 1: D beat E, C beats B, round two C faces D. If E is a liar, the second liar can't be B because if C and D are both knights, B is too. (for C to face B and D, C must beat D in round 1 or D is a liar and face B in round 2, and B must have one round 1 vs E making B's statement true.) So If E is a liar, the other liar is either C or D. But if it's D then D made the second round, but so did B (if he told the truth) and C (ditto). But all three can't be in the second round so E and D can't both be liars. So, if E is a liar then D and B are telling the truth (and hence C is a liar) and we have the scenarios in the paragraph above. (There are no others because if B beats E in round 1 instead of round 2 AND D loses in round one then C beats D and round two is C vs B making C's statement true, which is counter the 2-liar assumption.)

So, with two liars exactly, there are these possible tourneys:

Round 1: D beats E, C beats B; round 2: C vs D
Round 1: B beats D, E beats C; round 2: B beats E
Round 1: B beats C, E beats D; round 2: B beats E

These are the possibilities the speaker was faced with. Since knowing that a single joust took place was enough to determine the entire results of the tourney, that known joust must have been one that only appears once in the sets above, and in one of the possibilities with a round 2 winner. Only B vs D and E vs C fit that requirement, and both resolve to the same solution. So, we don't know which joust the speaker knew had occured, but we *are* sure that in round 1, knight B beat liar C and liar E beat knight D, and in round 2 B was trimphant over E.

Posted by Paul on 2005-08-16 03:15:14