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The Big Banquet (Posted on 2007-02-09) Difficulty: 3 of 5
A knight remembers a banquet that he had:

"Oh, that was a really nice banquet; all the liars and the knights of the kingdom were there, even the village fool. We all ate and drank and at the end of it the king asked each one of us to make a statement about someone else, I mean, to say if he is a liar or a knight. finaly, one statement was said about each one of us, and every one made the same statement except for the village fool who made a different statement. By the way, can you tell me how many people were in the banquet?"

"No", you say, "but I can tell you something else about the NUMBER of people that were in the banquet."

What can you tell? What statement did each one of the participants make?

See The Solution Submitted by Assaf    
Rating: 3.6667 (3 votes)

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Solution assuming unstated limitations | Comment 10 of 15 |
Assume the following:
1)  The fool is himself either a liar or a knight.
2)  Everyone made either the statement "He's a liar" or "He's a knight".
Assume everyone but the fool made the statement "he's a knight".
Then, as mentioned in my last email, every knight would need to point to a knight and every liar would need to point to a liar.  The graphs of the reference cycles would need to contain either all knights or all liars.  The problem is that the fool has nowhere to go.  If he's a liar, he goes in one of the liar's cycles, but then his statement "he's a liar" would have to refer to a knight, who can't be in the same cycle with liars.  If he's a knight, he goes in a knight's cycle but since he says "he's a liar" he can't be a knight.  Since the fool has nowhere to go, "he's a knight" doesn't work for the general statement.
Thus everyone but the fool says "he's a liar".  This means that each liar points to a knight and each knight points to a liar (other than the fools).  There can be any number of "closed reference cycles" containing an even number of liars and knights, but there must be at least one such cycle containing the fool.  What does this do to the cycle?
If the fool is a liar, then he is pointed to by a knight, and he points to a liar.  If the fool is a knight, then he is pointed to by a liar and he points to a knight.  Either way, the number of knights and liars not counting the fool is exactly equal.
In conclusion, there are an equal number of liars and knights not counting the fool, who could be either.  Everyone but the fool says "he's a liar", the fool says "he's a knight", and the number of people making the statements including the fool is an odd number.  That's all we know.

  Posted by AvalonXQ on 2007-02-11 04:31:53
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