MHB The Island of 10: Identifying the Knaves

  • Thread starter Thread starter castor28
  • Start date Start date
AI Thread Summary
To identify the 5 knaves among 10 inhabitants on the island, a visitor must ask a minimum of 8 yes-no questions. The reasoning is based on the fact that there are 252 possible combinations of knights and knaves, which exceeds 2^7, necessitating at least 8 questions to narrow down the options. A strategic approach involves using a "double negation" technique, where questions are framed to reveal the truth regardless of whether the respondent is a knight or a knave. By systematically eliminating half of the possible answers with each question, the visitor can efficiently determine the identities of the knaves. This method leverages the villagers' mutual knowledge to achieve the goal within the required number of questions.
castor28
Gold Member
MHB
Messages
255
Reaction score
0
On a fictional island there are 10 inhabitants, who all know each other, of which 5 are knights, who always tell the truth and the rest of them are knaves, who always lie.

A visitor to the island wants to determine the 5 knaves. What is the minimum number of yes-no questions he must ask the inhabitants in order to find the 5 knaves? (each question is asked to one person only).
 
Mathematics news on Phys.org
Here is the proposed solution.
[sp]
As there are $\binom{10}{5}=252>2^7$ possible answers, we need at least 8 questions.

There is an easy solution with 9 questions: ask 9 villagers something like "Are you a bird ?". The challenge is to find the answer in 8 questions.

There is a classical "double negation" trick that allows you to get the true answer to any yes/no question: if you ask "If I asked you <your question here>, what would you answer?", a liar will have to lie twice, and he will give you the correct answer.

The important clue here is that all the villagers know each other, and therefore each of them knows the answer.

You can make a list of the 252 possible answers, show that list to any villager, and ask "If I asked you if the correct answer is in the first half of the list, what would you answer?"

This will allow you to eliminate half of the list. You can then repeat the process with the other half. Since $252<2^8$, you will get the answer with at most 8 questions.
[/sp]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top