Fun with Math

Discussion in 'Off-topic' started by Z_Millla, Jul 6, 2014.

  1. Z_Millla

    Z_Millla Well-Known Member VIP Creative Architect

    Seeing as no one got the Blue Eyed Islander problem after a few days, I'm declaring myself the undisputed winner of this thread. Better luck next time.

    So let's talk about that final problem. I'll start with a hint that has a very good chance of revealing the answer as a whole:


    Not obvious enough? Here is an explanation of that hint:
    Hint Part 1:

    Hint Part 2:

    Hint Part 3:

    I'm guessing most can come up with the final reasoning at this point, but just in case not I'll finish the thought:


    GG
     
  2. Random0ne

    Random0ne Well-Known Member

    Why hadn't they all committed suicide by the time the foreigner came? All the foreigner said was that at least one islander had blue eyes, which they all would have been aware of already. By that logic, they should have been able to figure out their eye colors regardless of whether the foreigner visited or not.
     
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  3. Z_Millla

    Z_Millla Well-Known Member VIP Creative Architect

    Good question. I think I can give you an answer, but not sure if it will suffice.

    In the original scenario of 100 blue eyed people and 900 brown eyed people, it was known that there was at least one blue eyed person only because there was two or more blue eyed people. If a tribes man considered the scenario of a single blue eyed person, that lone individual would no longer have the information of "there is at least one blue eyed person".

    When the visitor says "there is at least one person with blue eyes", it creates a theoretical scenario where there could be a single blue eyed person who DOES know there is only one blue eyed person, unlike the previous scenario.

    So in short: There has to be a (at least theoretical) scenario in which there is a single blue eyed islander who is given the information "there is at least one person with blue eyes" to begin the redundant logic of figuring what happens in the 100:900 scenario.

    This is a very old problem and the logic is admittedly quite convoluted. If you want to try and find better explanations than mine just google "blue eyed islander problem".
     
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  4. Random0ne

    Random0ne Well-Known Member

    I'm still not entirely satisfied with the answer, but I'll take it :p
     
  5. Nelvs

    Nelvs New Member

    Seems like the question is actually used to demonstrate common knowledge http://en.wikipedia.org/wiki/Common_knowledge_(logic)

    An explanation I found on a math forum.

    n is the number of blue eyed person.

    n=1. Obvious

    n=2. Each blue-eyed person says: “If I am not blue-eyed, then there is one blue-eyed person on the island, and he will kill himself tomorrow because of what the traveller just told us.”

    “But then the next day comes and the other doesn’t kill himself! That must mean there are two blue-eyed people, and I am the other one!”

    n=3. Each blue-eyed person says: “If I am not blue-eyed, then there are two blue-eyed people on the island. They are each looking at each other and thinking (“If I am not blue-eyed, then there is one blue-eyed person on the island, and he will kill himself tomorrow because of what the traveller just told us.”). But neither will kill himself tomorrow, and they will realize their mistake tomorrow and kill themselves on the day after.”

    “But then the third day comes and they don’t kill themselves! That must mean there are three blue-eyed people, and I am the third!”
     
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  6. waterpower1

    waterpower1 Active Member

    Found a few extra questions online! Same as before, I'll restate the questions precisely as I was given them.

    Fun with Geometry
    Consider a pentagram (see picture below). There are five disjoint triangles initially. Just count the five ‘caps’ above the pentagon, do not count the larger triangles formed by using three of the five vertices, for those ‘triangles’ have lines going through them. By adding just two lines, you can go from five to 10 triangles. How?
    [​IMG]

    A Penny For Your Thoughts (hard)​
    You have 12 coins, one weighs slightly less or slightly more than the others. Using an equal arm balance and only making three weighings determine which one is different and whether it is slightly less or slightly more. Warning: the fact that you don’t know whether it is less or more is a major problem in the solution (it might help if you tried this one out with a picture)​

    You're Caught - Stop Your Whining (hard)​
    The King of Freedonia has caught 10 spies from the Kingdom of Sylvania who attempted to poison his wine. The King of Freedonia keeps 1000 bottles of wine in his cellar. The spies managed to poison one of the 1000 bottles but were caught before they could poison any more. The poison is a very special one that is deadly even at one-millionth the dilution of the poisoned wine bottle and takes EXACTLY 24 hours to kill the victim, producing no symptoms before death.

    The trouble is that the King doesn’t know which bottle has been poisoned and the wine is needed for the Royal Ball in exactly 24 hour’s time! Since the punishment for attempted regicide is death, the King decides to feed some of the wine to the spies. The King informs his wine steward that he will be able to identify the poisoned bottle by the time the ball starts and kill at most 10 of the spies.

    How does he do it?

    Legal 21 Officer,
    Young Saul, a budding mathematician and printer, is making himself a fake ID. He needs it to say he’s 21. The problem is he’s not using a computer, but rather he has some symbols he’s bought from the store, and that’s it. He has one 1, one 5, one 6, one 7, and an unlimited supply of + – * / (the operations addition, subtraction, multiplication and division). Using each number exactly once (but you can use any number of +, any number of -, …) how can he get 21 from 1, 5, 6, 7?

    Note: you can’t do things like 15+6 = 21. You have to use the four operations as ‘binary’ operations: ((1+5)*6)+7.

    Bridge Over Troubled Students
    There are four men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either one or two people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man

    Man 1: 1 minute to cross
    Man 2: 2 minutes to cross
    Man 3: 5 minutes to cross
    Man 4: 10 minutes to cross

    For example, if Man 1 and Man 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Man 4 returns with the flashlight, a total of 20 minutes have passed, and you have failed the mission.
     
    Last edited: Sep 15, 2014
  7. treyman77

    treyman77 Active Member

    waterpower1 does the king of Freedonia need all 1000 bottles of wine
    for the banquet?
     
  8. waterpower1

    waterpower1 Active Member

    Yes, although he can feed a few drops from each bottle to the prisoners. So for example he can feed a drop from the first 100 bottles to the first 1 prisoner, a drop from the second 100 bottles to the second prisoner, etc.
    Wrong set of questions :p Answers and hints are in the first set of comments
     
    Last edited: Sep 10, 2014
  9. Legend9468

    Legend9468 Well-Known Member VIP

    Well, if the spies were caught in the act of poisoning the bottles, then all you have to do is either look at the one which has already been disturbed, or whichever is at the start of the area (assuming they were trying to poison them all; plus that fact that people like to work in sequences).
     
  10. waterpower1

    waterpower1 Active Member

    Nope! The cellar is a round room with the entrance dead bang in the center of the room! Seriously though, no tricks.

    Hint
     
  11. TheTechGeek245

    TheTechGeek245 Well-Known Member VIP

    unless the king can have 900 bottles for the banquet i got nothing
     
  12. oiShocKWavesv

    oiShocKWavesv Well-Known Member VIP

     
  13. waterpower1

    waterpower1 Active Member

    That is honestly the best solution that I've seen anyone come up with, you've nailed that question perfectly :p

    I've added in an extra two problems for you guys to try. I've still got one or two (hard) ones left if you guys are interested.
     
    Last edited: Sep 12, 2014
  14. waterpower1

    waterpower1 Active Member

    Hints for the remaining ones:
    Fun with Geometry

    A Penny for your Thoughts

    Legal 21 Officer

    Bridge Over Troubled Students
     

    Attached Files:

    Last edited: Sep 15, 2014