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?

I normally stay clear of these logic puzzles, but I am procrastinating like mad and I am curious how well Mathematica can help me. I will try to fashion my computer-aided explanation as discussed in the forums.

First I created the set of all 24 possible tournament results; 3 initial joust set-ups times 4 possible outcomes for the first round times 2 possible outcomes for the final joust.

I wrote each of the men's statements as a function that can be applied to a single tournament and return a true or false according to whether or not the statement holds. When these functions are evaluated against all 24 tournaments, I sorted them and found: There are 6 tournaments for which all four men are liars, there are 12 tournaments where there are exactly 3 liars, and there are 6 tournaments with exactly 2 liars. This means that knowing the number of liars cannot determine a single tournament, but it does limit the number of liars/knights to three possibilities.

Now a little grunt work through seaching the truth table. First I group the 24 tournaments according to the number of liars; 4, 3, or 2. Also there are six possible pairings Oscar could have in mind (using initials): B-C, B-D, B-E, C-D, C-E, D-E.

All three groups have at least a pair of tournaments for each of these 6 possible pairings except the 2-liar group has only one tournament with a C-E joust. To be specific, the tournament where the Baron beats the Duke, the Earl beats the Count, and the Baron beat the Earl is the only torunament where the Count jousts the Earl and there are exactly two liars (the Count and the Earl). Furthermore, all of the collections of tournments (by number of liars) contain at least two tournaments with each pairing except for the given case.

Mathematica allowed me to list the possible tournaments, quickly compute the validity of each statement against the torunaments, and sort the tournaments with respect to the number of liars.

Add.:  I could have written a final function that counted the number of times each paring appeared in each group, flagging those that appear exactly once.  Why I didn't is a mystery to me :-)

Edited on August 15, 2005, 6:35 pm

Posted by owl on 2005-08-15 17:51:37