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Logic puzzle

  1. Oct 7, 2003 #1
    This one I read but is so good I wanted to post it reworded. Two groups of people live on this planet, pure liars and pure truth tellers. I went to a philosophy party(never mind that it's impossible) and met 3 people. The first says something but I don't catch it, the second one replies, "he said he was a liar." The third says to the second, "You are lying!" Is the third person a liar or truth teller?
    Last edited by a moderator: Oct 7, 2003
  2. jcsd
  3. Oct 7, 2003 #2
    Hello, jammieg!

    A good problem . . .
    Consider what the first person must have said.
    If he were a Truth Teller, he'd say, "I'm a Truth Teller."
    If he were a Liar, he'd lie and say, "I'm a Truth Teller."
    That is, NO ONE would ever say, "I'm a Liar."

    Hence, the second person lied.
    Therefore, the third person spoke the truth.


    A similar problem . . .

    An archaeologist discovered a temple with three gigantic statues
    guarding the entrance. To be allowed into the temple, he must
    identify each of the three gods.

    The God of Truth always tells the truth.
    The God of Falsehood always lies.
    The God of Diplomacy can tell the truth or lie.

    He asked the god on the left, "Who stands next to you?"
    "The God of Truth," was the reply.

    He asked the god in the middle, "Who are you?"
    "The God of Diplomacy," was the reply.

    He asked the god on the right, "Who stands next to you?"
    "The God of Falsehood," was the reply.

    The archeologist immediately identitied the three gods.
    Can you?
  4. Oct 7, 2003 #3


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    Truth is obviously not on the left, or he'd be lying about who is next to him.
    Truth is not in the middle, or he'd be lying about who he is.
    Truth is on the right, therefore, the god next to truth is falsehood, as stated by truth.
    That leaves only the left for diplomacy.

  5. Oct 7, 2003 #4


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    A man at a table is shot for having 53 bicycles. Why?

  6. Oct 7, 2003 #5
    He didn't pay for the parking!
  7. Oct 7, 2003 #6
    Ha! . . . Took me a few moments.
    The "Bicycles" (TM) are playing cards.
    (He probably had an Ace up his sleeve.)
  8. Oct 7, 2003 #7
    here's a simple one

    you have 9 coins one is counterfeit and a balance, using only two weighs, find the counterfeit coin (the counterfeit coin is lighter)
  9. Oct 8, 2003 #8


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    er how about finding the counterfeit coin without any weights at all? Simply try out all permutations of 4 coins on one side of the balance against 4 coins on the other side. Once you have found a permutation such that the two groups of 4 balance perfectly, you know that the unused coin is counterfeit.

    edit: or, for a more efficient solution, pick out any two groups of 4 and place them on each side of the balance. If they balance perfectly, then the unmeasured coin is counterfeit. If they do not balance, replace one coin on the lighter side with the unused coin, and continue replacing thus-far-unreplaced coins until you have a perfect balance. This takes at most 5 separate measurements.

    Actually, it technically only takes at most 4 measurements since if you have already replaced 3 of the coins on the lighter side of the balance, you can infer the 4th is counterfeit. Of course this only holds if you are correct in assuming that one of the coins is counterfeit in the first place. Reminds me of the superbowl episode of the Simpsons where Chief Wiggum is in jail and declares "one bar in a jail cell is always weak." He goes on to test all the bars except the leftmost one, and infers that it is the weak one, so he rans head-on into it only to fall flat on his back.
    Last edited: Oct 8, 2003
  10. Oct 9, 2003 #9


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    Weigh 3 coins on each side of the balance. If one side is lighter, the fake is in that group of 3. If not, it is in the set of 3 unweighed coins.

    Discard the 6 coins you know are true, and balance 2 of the remaining coins. If one is lighter, it is fake. If they weigh the same, the unweighed coin is fake.

    You could find one fake in 27 coins with 3 weighings, 1 in 81 with 4 weighings etc.

  11. Oct 13, 2003 #10
    I think there is one of those annyoing creative solution ways to find the fake coin in 2 measures, with 2 to 2000 or more coins, but I would rather throw it scales and all into the sea.
    Among all the various two typings of the people of this world there are those who are apt and preoccupied with finding and percieving the limitations of things and those who are always trying to find ways around those limits.
    Last edited by a moderator: Oct 14, 2003
  12. Oct 23, 2003 #11


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    Hmm... How's that exactly ?! [b(]
    I remmember doing this riddle quite some time ago and
    the top number I could solve it for was either 15 or 16
    for 3 weighings (I'm a bit tired right now so I won't
    check which one it was, for now). But 27 ?! No way !

    Peace and long life.
  13. Oct 23, 2003 #12


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    Oops... I'm sory Njorl, never mind. I was reffering to
    the riddle when you don't know weather the fake is lighter
    or heavier. If you do know it's either of these options
    it's another story of course.
  14. Nov 28, 2003 #13
    I have a variation on this idea that is equally as difficult. It goes along these lines:

    You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?
  15. Nov 29, 2003 #14
    Re: Re: Logic puzzle

    from which city are you?
    if it's a lier than he points you to the wrong way of his city which is the liers city, he points you to the truth tellers city because it's not his city and hes lier than it fits, now if it's truth teller than he points you to the truth tellers city because he tells than truth when asked.
    either way you go where the person tells you.
  16. Dec 1, 2003 #15
    There are 3 cannibals and 3 missionaries on one side of a river. If ever there are more cannibals than missionaries on one side of the river, the cannibals will eat the missionaries. The boat which will be used to go to the other side of the river seats 2, and at least one person must be in it in order for it to go from one side to the other. In what way can they travel to the other side of the river so that no missionaries are eaten?

    -------- ---------
    Last edited: Dec 1, 2003
  17. Dec 1, 2003 #16


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    Legend: | | = river, < > = boat, C = cannibal, M = missionary

    Code (Text):

              |    |<  > CCCMMM
              |<CM>|     CC MM
    C  M  <  >|    |     CC MM
    C         |< M>|     CC MM
    C         |    |<  > CC MMM
    C         |<CC>|        MMM
    CCC   <  >|    |        MMM
    CC        |<C >|        MMM
    CC        |    |<  > C  MMM
    CC        |<MM>|     C  M
    CC MM <  >|    |     C  M
    C  M      |<CM>|     C  M
    By symmetry, I've reduced the problem to one I already know how to solve, so I'm done.
    Last edited: Dec 1, 2003
  18. Dec 1, 2003 #17


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    Here's a good crossings problem:

    Chroot, Monique, Gale, and Hurkyl (four randomly chosen names) are fleeing from the law through a dark tunnel with only one flashlight. They are a good 17 minutes ahead of their pursuers. They find their path blocked by an underground ravine with only a long rickety bridge to permit them to cross. The bridge is surely too unstable to allow more than two of our heroes to cross at any one time, and it's too dangerous to attempt a crossing in the dark. Monique estimates she can cross the bridge in one minute, Chroot estimates that he can do it in two minutes. Gale thinks she needs five minutes to do it carefully, and Hurkyl needs to take ten minutes, just to be safe. Can these fantastic four cross the bridge before the law catches them?
  19. Dec 1, 2003 #18
    1 and 2 go across. 15 minutes left
    1 goes back. 14 minutes left
    5 and 10 go across. 4 minutes left
    2 goes back. 2 minutes left
    2 and 1 go across. done with no time to spare.

    Ok, now someone explain the one with the 3 people in a hotel.
    the one where they pay $10 each for the room, but the hotel clerk finds out the room was onlt $25, so he gives $5 to the bellboy to return. The bellboy can't figure out how to divide it evenly so he gives them $1 each and takes the other 2 for himself. They each are now out $9, which means they lost $27 in paying for the room, which leaves $3 left, but the bellboy only kept 2.
    Last edited: Dec 1, 2003
  20. Dec 1, 2003 #19
    how 'bout this logic puzzle?

    either God exists or God does not exist.

    which is it and provide proof.
  21. Dec 2, 2003 #20
    If you had one question and an all knowing robot will answer it for you but will kill you when it answers you what will you ask him? Assuming he was indestructble and can not be destroyed.
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