All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Another Race (Posted on 2003-12-15)
When I went to the race track in Racing Town, a town made up only of Knights which always tell the truth, Knaves which tell truths and lies in an alternating pattern, and Liars which always lie, a race between 6 citizens of that town had just finished.

I went to the 6 citizens and asked each of them the order that all 6 finished. They all gave me different responses, each thinking themselves as winning, displayed here left to right as first to last.

A: A C D E B F
B: B D F E C A
C: C D E F A B
D: D E F B A C
E: E B A D F C
F: F C B A E D

From what they said, I was able to figure out what the correct order was. What is it?

 See The Solution Submitted by Gamer Rating: 3.3636 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 3 of 9 |
The winner must be a knight or a knave as his first statement is true. Every odd numbered position he notes (counting first as #1) is true. The non-winners must be liars or knaves as their first statements are false. Every odd numbered position they note must be false.

Runners B and D agree on position 3. Since both can't be true, both must be false, and neither B nor D won.

Runners C and D agree on position 5. By the same logic, neither C nor D won.

That leaves 3 possible winners: A, E or F. The winner could be either a knight or a knave. If he's a knight, his ordering is correct. If he's a knave, there are two possible derangements (in this case, as it's 3 positions, cyclic permutations) of the three even positions, to falsify his even numbered statements. So there are three possible winners and three possible orders for each possible winner, so there are 9 possible orders to check out.

The 9 possible orders have to be checked out by seeing if the remaining speakers all can be evaluated as either liars or knaves.

If runner A is the winner and a knight then speaker B's only true statement would be about position 4, so runner A can't be a knight.

If runner A is the winner and a knave, the true order could be either AEDFBC or AFDCBE. In the former case, speaker C's only true statement would be about position 4 and so can be neither liar nor knave. However if the order is AFDCBE, then every other runner/speaker is a liar. So a solution is AFDCBE with A being a knave, and everyone else liars.

As far as uniqueness goes:
If E were the winner and a knight, the order would disagree with all speaker D's statements except the last. If E were a winner and the order was EDACFB, then speaker B's statements would be all wrong except the second position. If the order were ECABFD, then A's 2nd statement would be his only true one.

If the order were FCBAED, again speaker A's second statement would be his only true one. If the order were FABDEC, then speaker D's last statement would be his only true one, and if the order were FDBCEA, then speaker B's 2nd and 6th statements would be his only true ones.

 Posted by Charlie on 2003-12-15 09:22:05

 Search: Search body:
Forums (0)