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8 replies to this topic

### #1 bushindo

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Posted 22 November 2010 - 06:58 PM

Suppose that you encounter 6 people, each of whom is either a truth teller (always tells the truth), a liar (always tell lies), or a random answerer (always answer randomly with 'Yes' or 'No'). You know that in this group of 6, there are 3 random answerers, 1 truth teller, and 2 liars, but you don't know which person is which. The puzzle is to identify the 3 random answerers with the following information:

1) You can only ask yes/no questions to 1 person at a time
2) You can ask the same question of different people, or address different questions to the same person.
3) Everytime you ask a question, your question count increases by 1
4) Each person in the group of 6 knows what type of people the remaining 5 are.

So, determine a strategy that is guaranteed to identify the 3 random answerers. Bonus: what is the maximum number of questions required to guarantee correct identification?

Edited by bushindo, 22 November 2010 - 07:00 PM.

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### #2 tamjap

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Posted 22 November 2010 - 08:40 PM

Hypothetically, if a question is asked of the truth teller with an answer he cannot possibly determine accurately (e.g. "If asking a yes or no question of a random answerer, will his/her answer be correct"), does the truth teller's head explode?
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### #3 bushindo

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Posted 22 November 2010 - 10:16 PM

Hypothetically, if a question is asked of the truth teller with an answer he cannot possibly determine accurately (e.g. "If asking a yes or no question of a random answerer, will his/her answer be correct"), does the truth teller's head explode?

Yes, if that happens, the truth teller's head will explode. Actually, more seriously, I just realized that I made a mistake in constructing this puzzle. Turns out in the worst case I can narrow the 6C3 = 20 possible configurations down to 2, but no further. Therefore a solution is not possible, at least not for me. If a mod can delete this topic, I'd appreciate it very much. Sorry to those who spent time thinking about the problem as it was posted.
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### #4 araver

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Posted 23 November 2010 - 08:22 AM

Yes, if that happens, the truth teller's head will explode. Actually, more seriously, I just realized that I made a mistake in constructing this puzzle. Turns out in the worst case I can narrow the 6C3 = 20 possible configurations down to 2, but no further. Therefore a solution is not possible, at least not for me. If a mod can delete this topic, I'd appreciate it very much. Sorry to those who spent time thinking about the problem as it was posted.

Spoiler for

Seriously now
Spoiler for

Still not sure we need to delete this as we do not have a formal proof that it's impossible.

I propose we leave it as an open question for now or as "Bushindo's 3-Randoms-Hiding-Successfully Conjecture"

Edited by araver, 23 November 2010 - 08:25 AM.

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### #5 araver

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Posted 23 November 2010 - 09:28 AM

Actually, after thinking a little more ... I almost had a proof but it had a hidden fallacy in it. Just putting it here maybe it inspires an actual proof.

Spoiler for

OFF-TOPIC I admit I still enjoy this puzzle as I'm a twisted-version of a researcher
I enjoy mostly proving that something is either not possible or undecidable

Edited by araver, 23 November 2010 - 09:29 AM.

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### #6 araver

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Posted 23 November 2010 - 11:20 AM

And since I started ranting, let me continue with
Spoiler for

Edited by araver, 23 November 2010 - 11:23 AM.

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### #7 Lemeshianos

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Posted 23 November 2010 - 05:04 PM

I am just going to go for the Bonus question.
Since you are asking for the maximum and not the minimum, then my answer would be "Infinite number of questions"! (or just a number higher than the number anyone comes up with)
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### #8 wolfgang

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Posted 23 November 2010 - 07:36 PM

Hi...did anyone think about deviding the 6 people into two groups of 3?
In this case we will have only three possiblities:-
1- RRR......TLL
2- RRL......TLR
3- RLL......TRR
This method may help to find a possible solution,and minimize number of questions..
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### #9 bushindo

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Posted 23 November 2010 - 08:12 PM

And since I started ranting, let me continue with

Spoiler for

Good work, araver. You made some valuable contribution from an otherwise flawed puzzle. I'm glad you enjoyed it. I sure enjoy reading your proof.

Hi...did anyone think about deviding the 6 people into two groups of 3?
In this case we will have only three possiblities:-
1- RRR......TLL
2- RRL......TLR
3- RLL......TRR
This method may help to find a possible solution,and minimize number of questions..

This is a good start, but it is virtually impossible to distinguish between mirror pairs. Suppose we narrow the problem to either possibility 2 or 3 in your example, it is virtually impossible to determine which is the correct one from there.

Edited by bushindo, 23 November 2010 - 08:13 PM.

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