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A Dinner Puzzle

  1. Feb 18, 2007 #1
    We know that Truthers always tell the truth but Liars will always speak falsely.

    An island comprises entirely of Liars and Truthers..

    Precisely 45 people attended a dinner in the island. Each one of them sat at a big round table.

    After the dinner was over, each of the attendees was asked about their neighbours, and each stated that they were seated between one Liar and one Truther.

    It transpired later that precisely two of the Truthers were mistaken in their statements.

    Determine the number of Liars And Truthers attending the dinner.
    Last edited: Feb 18, 2007
  2. jcsd
  3. Feb 18, 2007 #2
    Sorry, my first answer was wrong:

    K Sengupta

    It looks quite possible that there was only two Truthers,
    that they were mistaken,
    and that they were not sitting close to each other.

    I am sure if there is only one solution to this problem.

    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3

    I tried for 9 people, and found 6 truther and 3 liars.
    For more people I don't know.

  5. Feb 18, 2007 #4
    Deleted it sorry my first answer was rubbish :smile: kinda misread the question thus it was stupid.


    I have it I think, 15 liars and 30 truthers.

    Pattern starts LTLTTT and then LTTLTT til the end.

    I think that works? Two mistaken truthers at the start.
    Last edited: Feb 19, 2007
  6. Feb 18, 2007 #5
    I think that there might be 14 liars and 31 truth tellers.I think this is the answer.
    Last edited by a moderator: Feb 19, 2007
  7. Feb 19, 2007 #6

    ten of the best characters
    Last edited: Feb 19, 2007
  8. Feb 26, 2007 #7
    A solution

    T=truth teller
    L= lier
    T'=mistaken truth teller

    It is assumed that T and L both see T' as a T (they don't know of mistake)

    possible adjacent seating for mistaken truth teller T' are
    LT'L and TT'T

    all other seats must be LTT LTT LTT LTT LTT etc

    so a solution is LTT ..LT'L TT'T ..LTT LTT LTT LTT UNTIL 45 SEATS
    so this gives 15 L and 30 T with two of the Ts being T'
  9. Feb 27, 2007 #8
    Vamfun don't forget to use the spoiler colour see post no.6 above.



    Well that's two of us with that solution, I think it must be the right one, if you try other combinations it doesn't work, so I'm hazarding a guess it's the only solution, although?
    Last edited: Feb 27, 2007
  10. Feb 27, 2007 #9
    Yep! I'll see if I can prove it:

    For starters, we must state a couple assumptions:
    1) The liars all told falsehoods (NONE of the liars were next to both a truth teller and a liar)
    2) Exactly two of the truth tellers were mistaken, the rest were correct in stating that they sat next to both a liar and a truth teller.

    Now, there's *one* possibility that's really easy, which is:


    But because we know that there are at least two truth tellers who attended the dinner, this answer is wrong.

    We can also easily deduce that no two liars sat next to each other. The possibilities would be as follows:

    ...TFFT... <-- this would make both middle liars correct, and is disqualified
    ...TFFF... <-- this would make the leftmost liar correct, disqualified
    ...FFFT... <-- this would make the rightmost liar correct, disqualified
    ...FFFF... <-- this is plausible, but leads ONLY to Solution A, which is wrong.

    Therefore, no two liars sat next to each other.

    From that, we know that there are at MOST 22 liars, which means at LEAST 23 truth tellers. And that means that there are at least 21 truth tellers who were correct in their statements. As a result, we know that there are a minimum of 11 liars, because the 21 truth tellers all correctly reference at least one liar who was known to be at the dinner, but half of the accurate truth tellers may have been referencing the same liars.

    So! Between 11 and 22 liars, proven.

    We've shown that no liars can be seated next to each other. This means that *EVERY* liar must be seated in between TWO truth tellers. And, the vast majority of those truth tellers must in turn be seated next to *another* truth teller because the majority of truth tellers are correct in their assesments.

    The only thing left are the anomolies. The anomolous truth tellers can either be seated as TTT or FTF. The remainder of people MUST be situated in the pattern: ...TTFTTFTTF... The pattern itself is therefore a multiple of 3.

    Hence, we are left with possibilities:
    insert a TTT and a TTT
    insert a TTT and a FTF
    insert a FTF and a FTF

    But notice that by inserting a TTT into the pattern, we must NECESSARILY be adding *one* person (a truth teller) to a single phase of the pattern. And conversely, by inserting an FTF into the pattern, we must necessarily be REMOVING one person (a truth teller) from the pattern. So, let's reanalyze our possibilities:

    insert a TTT and a TTT - wrong, this would add 2, and our total would not be a multiple of 3.

    insert a TTT and a FTF - possible

    insert a FTF and a FTF - wrong, this would remove 2, and our total would not be a multiple of 3.

    As a result, we must take our base pattern of TTFTTFTTF, and insert a TTT and an FTF, in order to keep the total at 45. Where we add them is irrelevant.

    If we start with our base pattern, we get that there must be twice as many truth tellers as liars, or 30 truth tellers and 15 liars. Then, we put in the necessary TTT, and we get 31 truth tellers and 15 liars. Finally, we put in the necessary FTF, and we go back to having 30 truth tellers and 15 liars.

    Therefore, the solution of 30 truth tellers and 15 liars is unique.


  11. Feb 27, 2007 #10
    Superb. Well done :smile: I can't see a flaw in your logic.:smile:
  12. Feb 27, 2007 #11
    Thanks.... not sure how to do it but here goes.
    Last edited: Feb 27, 2007
  13. Feb 28, 2007 #12
    That's it, if you can edit your last post with the answer with this colour so it's spoilered. I'm not sure, but I suspect you could ask a mentor or admin to do that if not?
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