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Thread: Mind Trap's

  1. #61
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    Re: Mind Trap's

    Quote Originally Posted by GHERKIN
    Haha, no. It's all logic. It is not a trick
    It isn't a trick if one assumes they are catholic monks
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

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    Re: Mind Trap's

    Okay, but they aren't. I have provided all relevant details, and specifically noted that logic alone solves the problem. No additional assumptions are used, save that the monks are perfectly logical, and when logic comes into play, they do not fail. You just have to go through all of the logical possibilities to get the answer.
    He who has an ear, let them hear.

  3. #63
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    Re: Mind Trap's

    No takers?

    They leave in as many days as there are monks with marks.

    If there was one monk with a mark, he would know it on the original night (night 0), as no other monk would have a mark.

    If 2 had marks, they would know on the first (additional) night, because the monk they might have otherwise thought was the only one with a mark had not left - implying that other monks had marks.

    If 3 monks had marks, they would know on the second (additional) night, because the monks they would have otherwise thought were the only ones with marks had not left on the second day - implying that other monks (them) had marks.

    Note that in each case, the monks would understand the answer on (#monks-1) night (after the announcement), and would leave on (#monks) day.

    This pattern continues all the way to 100.


    Here's another:

    Suppose you have 12 coins, each of which is identical, except one which is either slightly heavier or slightly lighter - you don't know which.

    You have a scale, of the old fashioned variety, which can measure two weights against each other to show which is heavier, like lady justice might carry.

    In only 3 weighings, how can you determine which coin is heavier or lighter, and if it is heavier or lighter. You cannot tell the difference by feel alone, and again there is no trick in this question - it is only logic.
    He who has an ear, let them hear.

  4. #64
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    Re: Mind Trap's

    Wouldn't this also cause monks that did not have marks to leave as well? Say there were only 3 monks, and 2 had marks. The first monk with a mark waits for the second monk with a mark to leave/realize they have a mark as well. The fact that both monks with marks are still there, causes the 3rd monk to wrongly assume they are waiting on him. So he leaves.


    answer[
    put 6 coins on each side, then switch 3, then, remove all coins but the 3 that causes the unbalance and weigh 2 of them. If the 2 on the scale or equal, then the on left over is the unbalanced coin.
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  5. #65
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    Re: Mind Trap's

    Quote Originally Posted by Gherkin View Post
    Suppose you have 12 coins, each of which is identical, except one which is either slightly heavier or slightly lighter - you don't know which.

    You have a scale, of the old fashioned variety, which can measure two weights against each other to show which is heavier, like lady justice might carry.

    In only 3 weighings, how can you determine which coin is heavier or lighter, and if it is heavier or lighter.
    It can as easily be done with 14 coins if one of them is known not to be the odd one.

    ---------- Post added at 03:32 AM ---------- Previous post was at 02:54 AM ----------

    Quote Originally Posted by Gherkin View Post
    "I have noticed that some of the monks in this monastery have developed marks upon their foreheads. If any monk determines that they have a mark upon their forehead, they must leave the monastery by sunrise of the day following their discovery."
    [emphasis added]
    There is an implicit assumption in this riddle that the monks who have developed forehead marks are unaware that they've done so, and there is no logical reason presented for any of them to assume that he won't develop such a mark between the time of the announcement and the morning of departure. So, while the monks who leave will all have reasoned that they must have had a mark at the time of the head monk's announcement, the ones who remain cannot be certain that they lack such a mark.

  6. #66
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    Re: Mind Trap's

    Gherkin, I would have been able to answer your question had you told me there is no need for a specific answer (i.e. six monks, two days).

    ---------- Post added at 05:41 PM ---------- Previous post was at 04:53 PM ----------

    Quote Originally Posted by Gherkin View Post
    No takers?


    Here's another:

    Suppose you have 12 coins, each of which is identical, except one which is either slightly heavier or slightly lighter - you don't know which.

    You have a scale, of the old fashioned variety, which can measure two weights against each other to show which is heavier, like lady justice might carry.

    In only 3 weighings, how can you determine which coin is heavier or lighter, and if it is heavier or lighter. You cannot tell the difference by feel alone, and again there is no trick in this question - it is only logic.
    Weigh six of them against the other six, one side will way more - discard the lighter side.
    Split the nest group of six into two groips of three, weigh them, and do the same asabove.
    You will have three coins, put one aside.

    Weigh the two coins against each other. If one ways more you know which it is. If they weigh the same then the coin set aside is it.



    Edit: nvm I just realized you didnt state if it was specifically lighter or heavier, otherwise that was really easy.

  7. #67
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    Re: Mind Trap's

    Quote Originally Posted by Gherkin
    Here's another:

    Suppose you have 12 coins, each of which is identical, except one which is either slightly heavier or slightly lighter - you don't know which.

    You have a scale, of the old fashioned variety, which can measure two weights against each other to show which is heavier, like lady justice might carry.

    In only 3 weighings, how can you determine which coin is heavier or lighter, and if it is heavier or lighter. You cannot tell the difference by feel alone, and again there is no trick in this question - it is only logic.
    I'm at a loss.

    Quote Originally Posted by Galendir
    It can as easily be done with 14 coins if one of them is known not to be the odd one.
    Is this any different than 13 coins?
    Udabindu yathāpi pokkhare
    Padume vāri yathā na lippati,
    Evaṃ muni no palippati
    Yadidaṃ diṭṭhasutaṃ mutesu vā.

  8. #68
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    Re: Mind Trap's

    Question to opponent.When the day after tomorrow is yesterday, today will be as far from Wednesday as today was from Wednesday when the day before yesterday was tomorrow. What is the day after this?
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  9. #69
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    Re: Mind Trap's

    Quote Originally Posted by MindTrap028 View Post
    When the day after tomorrow is yesterday, today will be as far from Wednesday as today was from Wednesday when the day before yesterday was tomorrow. What is the day after this?
    ~ Friday ~


    ---------- Post added at 02:26 AM ---------- Previous post was at 02:25 AM ----------

    Quote Originally Posted by Soren View Post
    Is this any different than 13 coins?
    Yes.

  10. #70
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    Re: Mind Trap's

    Incorrect Galendir.
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  11. #71
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    Re: Mind Trap's

    Quote Originally Posted by MindTrap028 View Post
    Incorrect Galendir.
    "...what is the day after this?"
    Oops! Missed this part. That would be
    ~ Saturday ~

  12. #72
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    Re: Mind Trap's

    Sorry, wrong again.
    Thursday, is always the day ofter wednesday
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  13. #73
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    Re: Mind Trap's

    Quote Originally Posted by MindTrap028 View Post
    Wouldn't this also cause monks that did not have marks to leave as well? Say there were only 3 monks, and 2 had marks. The first monk with a mark waits for the second monk with a mark to leave/realize they have a mark as well. The fact that both monks with marks are still there, causes the 3rd monk to wrongly assume they are waiting on him. So he leaves.
    No. they have no ability to make logical errors.

    put 6 coins on each side, then switch 3, then, remove all coins but the 3 that causes the unbalance and weigh 2 of them. If the 2 on the scale or equal, then the on left over is the unbalanced coin.
    No. I don't see how your first step helps anything, how switching 3 helps, how you figure the 3 that cause the unbalance, and your last step does not account for if the 2 are unbalanced (2/3 chance).

    Quote Originally Posted by Galendir
    There is an implicit assumption in this riddle that the monks who have developed forehead marks are unaware that they've done so, and there is no logical reason presented for any of them to assume that he won't develop such a mark between the time of the announcement and the morning of departure. So, while the monks who leave will all have reasoned that they must have had a mark at the time of the head monk's announcement, the ones who remain cannot be certain that they lack such a mark.
    Great point. In the future I will have to note this as well.

    Quote Originally Posted by soren
    nvm I just realized you didnt state if it was specifically lighter or heavier, otherwise that was really easy.
    Yes.

    The first step is not to measure 6 against 6 (it would always give no new information), and even after you have ruled out which coin it is, you cannot discount those coins, but must still use some of them.
    He who has an ear, let them hear.

  14. #74
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    Re: Mind Trap's

    Quote Originally Posted by GHERKIN
    No. they have no ability to make logical errors.
    It is not a logical error, it is factual error. It is a logically sound conclusion.

    If however they are simply unable to be wrong.. then they would all (who had the mark)leave on the first day. because if they thought they had the mark, they would inherently be correct.
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  15. #75
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    Re: Mind Trap's

    It is not a logical error, it is factual error. It is a logically sound conclusion.
    Let me show you how this is a logic error:

    If there were 3 monks, and monk #3 did not have a mark, but the other two did - he could only leave if he misassumed that he did have a mark. Here is how the timeline would break down with only correct logical deductions.

    ======

    Night 0 (night of the announcement) -

    Monk 1 sees monk 2 with a mark, and monk 3 without one.
    Monk 2 sees monk 1 with a mark, and monk 3 without one.
    Monk 3 sees monks 1 and 2 with marks.

    Monk 1 on this night would not assume that he had a mark, due to incomplete information, and the knowledge that the minimum provisions of the announcement had been made (at least one monk had a mark), without it necessarily being him.

    Monk 2 would do the same.

    Monk 3 would do the same.

    No monks would leave on Day 1.

    ======

    Night 1

    Monk 1 sees that monk 2 has not left, yet monk 3 has no mark. This necessarily means that other monks have marks, and as he is the only remaining monk, he must have one.

    Monk 2 sees that monk 1 has not left, yet monk 3 has no mark. This necessarily means that other monks have marks, and as he is the only remaining monk, he must have one.

    Monk 3 sees that monks 1 and 2 have not left, and correctly deduces that enough information had not been available to the infallibly logical monks on night 0. He further realizes that if the monks (1 and 2) do not leave tomorrow due to the information being complete enough to correctly make a deduction from, that he will have a mark as well - which will become apparrent the next day. (In this case, all three would have the same circumstance from their perspectives [all 3 with marks], and they would all leave Day 3, after discovery of this fact on Night 2.)

    Monks 1 and 2 would leave on Day 2.

    ======

    The monks never wrongly assume, and never fail to make correct deductions from complete information. Their logic is flawless.

    ======

    Nobody want to try the coin problem again?
    He who has an ear, let them hear.

  16. #76
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    Re: Mind Trap's

    Quote Originally Posted by GHERKIN
    The monks never wrongly assume, and never fail to make correct deductions from complete information. Their logic is flawless.
    Then they would all leave on day 1. (or asap)

    Because in order to stay to day 2, they must assume they do not have a mark. If they can not be incorrect, then they will leave.

    Quote Originally Posted by GHERKIN
    If there were 3 monks, and monk #3 did not have a mark, but the other two did - he could only leave if he misassumed that he did have a mark
    But the valid logical conclusion is that as long as the two monks he sees have marks are still there, that he has a mark as well. In short, in the same moment the second monk correctly assumes that he has a mark. The 3rd monk would logically valid, but factually wrong assume he has a mark as well. If the logic is valid for the second monk, then it is valid for the 3rd monk as well.

    Just because the answer is ultimately wrong, doesn't mean it is illogical.
    A wrong assumption, is not a logical fallacy. It is a factual fallacy.

    Quote Originally Posted by FALLACYS
    A fallacy is, very generally, an error in reasoning. This differs from a factual error, which is simply being wrong about the facts.
    http://www.nizkor.org/features/fallacies/
    Again, if they are incapable of making a wrong assumption, then they are all required to make an assumption the first day.

    As they can not be incorrect, all with a mark would correctly leave.. though without any logical reason.
    I apologize to anyone waiting on a response from me. I am experiencing a time warp, suddenly their are not enough hours in a day. As soon as I find a replacement part to my flux capacitor regulator, time should resume it's normal flow.

  17. #77
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    Re: Mind Trap's

    Quote Originally Posted by MindTrap028 View Post
    Sorry, wrong again.
    The pronoun "this" is ambiguous. It can grammatically correctly be understood to either refer to 'Wednesday' or 'today'. This ambiguity is necessary to allow your preferred answer to be considered correct. But, by the same token it also permits the other interpretation to be correct (or at least, not "wrong".)
    Is the riddle merely a ploy at misdirection or is it a legitimate logical challenge? It is difficult to justifiably insist that it must be one and not the other.
    One might wish to argue that had it been solely a logical challenge, it wouldn't have been ambiguously worded. But, this assumes that the person presenting the riddle is rather careful about his wording, and this, I think you'll humbly admit, is an unwarranted assumption.

    ---------- Post added at 04:20 AM ---------- Previous post was at 02:34 AM ----------

    Quote Originally Posted by Gherkin View Post
    Nobody want to try the coin problem again?
    Too easy. http://www.onlinedebate.net/forums/s...s-(challenging)

    Quote Originally Posted by MindTrap028 View Post
    Then they would all leave on day 1. (or asap)

    Because in order to stay to day 2, they must assume they do not have a mark. If they can not be incorrect, then they will leave.
    They do not assume that they do not have a mark. Neither do they assume that they do have a mark. They don't assume anything (at least, not in the way you mean.) They deduce. They cannot (yet) deduce whether they in fact have a mark or not.

    But the valid logical conclusion is that as long as the two monks he sees have marks are still there, that he has a mark as well.
    No. This is entirely invalid. There is no logical justification for him to reason that he, too, has a mark.

    In short, in the same moment the second monk correctly assumes that he has a mark. The 3rd monk would logically valid, but factually wrong assume he has a mark as well. If the logic is valid for the second monk, then it is valid for the 3rd monk as well.
    No, it isn't. The second monk sees one other monk with a mark. The third monk sees two other monks with marks. They have differing information from which to draw their inferences.
    Suppose only monk 1 had a mark. Monk 1 would see no marks on any other monk and so immediately deduce that he was the sole possessor of one and leave the following morning. None of the other monks could reason this way and would not immediately know whether he too had a mark, so he would not leave in the morning. If instead monk 2 also had a mark, then monk 1 would not have been able to deduce that first evening that he alone had a mark, and so would not have left in the morning. Seeing that no monk left that first dawn, each would deduce that at least two monks had marks. Since monks 1 & 2 each could see only one mark, each would correctly deduce that he, himself had the other mark, and leave on the second morning, but no other monk could make this deduction since they all can see both marks.

    Just because the answer is ultimately wrong, doesn't mean it is illogical.
    A wrong assumption, is not a logical fallacy. It is a factual fallacy.
    The monks don't assume; they deduce. They can only reach a false logical conclusion if they reason from a false premise, and as they make no false assumptions, their premises are true and their conclusions are sound.

    Again, if they are incapable of making a wrong assumption, then they are all required to make an assumption the first day.
    This is utterly mistaken. They need make no assumptions at all.

    As they can not be incorrect, all with a mark would correctly leave.. though without any logical reason.
    They don't magically know the answer to any question. What they do is reason, and reason perfectly, that is, with flawless logic.

  18. #78
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    Re: Mind Trap's

    Quote Originally Posted by Gherkin View Post
    No takers?

    They leave in as many days as there are monks with marks.

    If there was one monk with a mark, he would know it on the original night (night 0), as no other monk would have a mark.

    If 2 had marks, they would know on the first (additional) night, because the monk they might have otherwise thought was the only one with a mark had not left - implying that other monks had marks.

    If 3 monks had marks, they would know on the second (additional) night, because the monks they would have otherwise thought were the only ones with marks had not left on the second day - implying that other monks (them) had marks.

    Note that in each case, the monks would understand the answer on (#monks-1) night (after the announcement), and would leave on (#monks) day.

    This pattern continues all the way to 100.
    There is a fatal logical flaw. Who is the first to leave? Not by name but by rationale. If all the other monks are without marks and the set knows it to be true that someONE in the set is marked then the one marked can deduce that he must be marked. If there are any more than one then all will conclude it is the other and no one will leave.

    ---------- Post added at 01:47 PM ---------- Previous post was at 01:07 PM ----------

    Upon further thought what matters is that the quantity of marked monks is known. With that known all can observe enough information to come to a conclusion. Without knowing how many no one has enough information.

    In my example of a single mark, the quantity was inherently known to be one, without needing to specify. Quantities greater than one do not carry the same inherent value of quantity.

    Additionally how are the monks to know when all have left? By memory the head monk did not specify he would conclude the process. (or did he?)

    ---------- Post added at 02:07 PM ---------- Previous post was at 01:47 PM ----------

    Regarding the 12 coins:

    IF the heavier/lighter is known, I have the process. Either way does not matter as long as it is known. Without knowing I can only resolve it with a fourth measurement.

    Twelve is such a beautiful number (I wish the world operated on base 12), it is evenly divisible into so many parts.
    1. Divide the coins into Groups A, B, and C each with 4 coins.
    2. Compare A and B.
    - If they are identical it must be C that is unique.
    - If they are different the known unique (heavier/lighter) is the correct pile.
    - If heavier/lighter is unknown step 2b is necessary.
    2b. Compare B and C.
    - If A and B were identical then they set Baseline for weight. With B as Baseline C is unique and known to be heavier/lighter.
    - If A and B were different Baseline is unknown until comparing to C. A or B will be identical to C and they set Baseline for weight and the unique Group (A or B) is now known as heavier/lighter.
    3. Name the four coins from the unique pile W, X, Y, Z.
    4. Compare W+X to Y+Z.
    - Heavier/Lighter has been determined previously, select the Unique Pair.
    5. Compare W(or Y) to X(or Z).
    - Heavier/Lighter has been determined previously, select the Unique Coin.


    -----

    The day after:

    Thursday


    -----

    Magician and the bridge:
    I can juggle. I can engineer a bridge. I can calculate Live Load (capacity) and Dead Load (structure).

    Use all the math you like, some of which made it sound impossible to juggle at all. The timing is the critical factor. In the process of juggling there is never a single moment where the Magician exerts the critical weight on the bridge. Think about a really low juggle, then think of a really high juggle. In the really high juggle it can easily be seen that at any given time two objects are in the air and a maximum of one is in the hand.
    Last edited by Mehkael; July 14th, 2011 at 10:24 AM. Reason: clarity
    All men SHOULD have equal opportunity to live their life to the fullest. Be as happy, productive, carefree, and content as possible. The trouble is we are NOT created equal.

  19. #79
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    Re: Mind Trap's

    Quote Originally Posted by Mehkael View Post
    There is a fatal logical flaw. Who is the first to leave?
    I think you are confused. There is no "first" to leave. All the monks who leave, leave on the same morning at dawn.
    If all the other monks are without marks and the set knows it to be true that someONE in the set is marked then the one marked can deduce that he must be marked. If there are any more than one then all will conclude it is the other and no one will leave.
    What do you mean "it is the other"? It is true that no one will leave the first dawn following the announcement if there are more than one marked monk. It is because of this fact that each of the two monks with marks can reason that there must be another monk besides the one he sees that is also marked. Since neither can see a second monk with a mark (assuming exactly two monks are marked), he rightly deduces that he himself is the second marked monk, and they both leave the second dawn. All the non-marked monks see exactly two other monks with marks and expect them both to leave the second dawn if there are no more marked monks. But suppose you are one of the other monks, and observe that the two monks with marks that you can see do not leave the second dawn as expected. You would deduce that they, too, must each see two other monks with marks. Each of you would then deduce that you must be the third monk with a mark, and you would all leave together the third dawn. Etc., etc.
    Upon further thought what matters is that the quantity of marked monks is known. With that known all can observe enough information to come to a conclusion. Without knowing how many no one has enough information.
    This is only partly right. If the quantity is known, then it is very simple for the monks to count the observed marks and know whether or not they have the last and all would leave the morning following the announcement. This wouldn't be a challenge for anyone who can count.
    Without knowing the quantity of marked monks, it is true that there is not enough information yet to know whether one carries a mark. But with each passing morning, the information increases until every monk with a mark will be able do deduce that he has one, and all leave the next day.

    Additionally how are the monks to know when all have left? By memory the head monk did not specify he would conclude the process. (or did he?)
    Yes, you are confused. The monks do not leave one by one on successive days. They leave together the same morning.

    Regarding the 12 coins:
    IF the heavier/lighter is known, I have the process.
    If this is known beforehand, the coin can be identified in three weighings out of a pile as large as 27.

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    Re: Mind Trap's

    Quote Originally Posted by Galendir
    What do you mean "it is the other"? It is true that no one will leave the first dawn following the announcement if there are more than one marked monk. It is because of this fact that each of the two monks with marks can reason that there must be another monk besides the one he sees that is also marked. Since neither can see a second monk with a mark (assuming exactly two monks are marked), he rightly deduces that he himself is the second marked monk, and they both leave the second dawn. All the non-marked monks see exactly two other monks with marks and expect them both to leave the second dawn if there are no more marked monks. But suppose you are one of the other monks, and observe that the two monks with marks that you can see do not leave the second dawn as expected. You would deduce that they, too, must each see two other monks with marks. Each of you would then deduce that you must be the third monk with a mark, and you would all leave together the third dawn. Etc., etc.
    Can you actually prove that this pattern holds through 100? I can see informally why it would make sense that it does--and I certainly think that it does continue--but to make the proof rigorous you would need to demonstrate it (proof by induction might work well).
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