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 Liars can figure (Posted on 2017-04-24)
The annual round-table meeting pf the L&K Island's government meeting requires quorum of 5 members.
(i). Assuming that there are n members present and each one declares:
"I'm sitting between 2 Liars."
Can you estimate how many of them are Liars?

Another Liar arrives and takes a seat between 2 members and declares:
"I'm sitting between 2 Liars."
Apparently he is coherent (i.e. he lies) no matter where he placed himself.

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 I am between two answers (spoiler) | Comment 1 of 4
PART I

Two knights cannot be sitting together.  And three liars cannot be sitting together.  Therefore, going around the table, we must have a knight, 1 or 2 liars, a knight, 1 or 2 liars, etc.

This means that the number of liars must be between n/2 (if there is exactly one liar per knight) and 2n/3 (if there are exactly two liars per knight).  Final answer.

Note that for n < 6, this means that the number of liars are known exactly.  For instance, when n = 5, the number of liars are between 2.5 and 3.33,  In other words, there are exactly 3 liars.

PART II

As asked, there is no extra information if a liar sits down and says that he is seated between two liars.  All this means is that he sat next to a knight, which is always possible.  Therefore, the part (1) answer is unaffected.

The slightly more interesting question, which was not asked, is what if everybody at the table still says that that they are seated between two liars, even after the new liar is seated.  In this case, this means that there was originally at least one liar who was seated between two knights.
This new information reduces the high end of the range, which becomes 1+ 2(n-2)/3, or (2n-1)/3.  But, as I have said, this is the answer to a question that was not asked.

 Posted by Steve Herman on 2017-04-24 10:13:37

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