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Everything posted by harey

  1. There is a hotel with an infinite number of rooms, all rooms occupied by little green men (one man in a room). An infinity of little blue man arrive and each one asks for a room. No problem, the manager moves the blue man from the room n to the room 2*n, freeing the odd-numbered rooms for the green men. So far, loosely copied from https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel. It turns out that the blue men sing between noon and midnight and sleep between midnight and noon while the green men sleep between noon and midnight and sing between midnight and noon. Complaints. The manager decides to group them. Conveniently, the rooms are in a straight line, numbered from left to right. While there is a green man left to a blue man, he makes them change the rooms. Eventually, all the blue men leave. - how many rooms are free? - how many rooms remain occupied? - what is the number of the first occupied room?
  2. A big, big, big hotel

    Brilliant, surprisingly interesting and leading to a major annoyance. Well, now back to the original problem. Everybody leaves, the hotel is empty. The green and blue men arrive all together and go to the room they occupied previously. Is this problem equivalent to the first one? Precision: By equivalent, I mean that the problems are announced in a slightly different way, but the answers remain the same. Something this way: Aunt: If I give you 5 bananas and take back 2 bananas, how many bananas have you? John: I do not know. Aunt: I met your teacher and she told me you make this kind of problems at school. John: Yes, but we do it with apples. Just do not tell me that the men know now the room numbers. Besides being clairvoyant, they can move in time. Forwards, backwards and sideways.
  3. When midnight strikes

    Sorry for 'discard', I meant these numbers build a {set} from which elements can be removed. And yes, we remain in rationals. What if l only paint those between 0.5 and 0.6, even leaving .53745 unpainted? Did I not paint an infinite number of numbers? We turn in round here. You claim that if you can remove an infinite amount of elements from a set, the set will be empty - I claim there may remain some elements (and even more then removed, depending on circumstances). I suppose you apply a 'law' or a 'theorem'. Can you give me the (Wikipedia) reference?
  4. When midnight strikes

    Maybe an idea. There is an infinite number of numbers between 0 and 1. I have the opportunity to discard an infinite number of numbers between 0 and 1. How much are you willing to bet that nothing remains between 0 and 1?
  5. When midnight strikes

    If a coin can be discarded, it does not mean it will be discarded. I would reformulate it: The key question is this: will all coins that are kept at a certain event ever be discarded at a later event? BTW, we can establish a bijection between Al's and Bert's coins. The coins bear green numbers. After each step, Al renumbers them and assigns them blue numbers 1, 3, 5, ... His blue numbers will match the (green) numbers in Bert's box. For any number of steps. I think it is legal to assume it is true even for N-> inf. Another way to prove that Bert's box will not be empty: graphical presentation. The number of coins in his box is a straight line (at 45 degrees). How is that it suddenly drops to 0? And maybe a corollary: Bert never discards more coins that he receives. How is that when he has, let's say 8 coins, he can have less in a later stage? If we reason with {coins} and {events}, don't you see a 1-1 relation?
  6. What Question Must Be Asked?

    I was so stuck in my solution that is exactly the same as yours except that the exclusion came earlier that I reacted too quickly. Next time, I will read more carefully, promised
  7. When midnight strikes

    After N steps, they will have received 2*N coins and withdrawn N coins. At that moment, there will be 2*N-N coins in the box. If the box is empty at midnight, this implies: limit(2*N-N)(for N->inf) = 0 At least a little bit surprising. @ThunderCloud I have some troubles to refute your argument. If you remove an infinity of finite numbers from infinity of finite numbers, it does not imply no finite number remain. (Not sure I am convincing and clear enough.) Counterargument: Al removed all coins 1 - N, coins > N remain. If N -> inf, numbering looses it's sense, but he did not remove all coins. As for Charlie, I am ruminating, too. The first idea: every number will remain with p=1/2. Wrong, 1 will be more likely removed than 99. 2nd idea: 1st step, 2 coins: p(removing 1)=1/2 2nd step, 3 coins: p(removing 1)=p(1 was not removed in the first step) * 1/3 = 1/2 * 1/3 = 1/6, p(1 remaining after 2nd step)=1 - 1/2 - 1/6 = 1/3 3rd step, 4 coins: p(removing 1)=p(1 not yet removed) * 1/4 = 1/3 * 1/4 = 1/12, p(1 remaining after 3rd step)=1 - 1/2 - 1/6 - 1/12 = I will not venture further, but this will not converge to 0. (Compare to 1 - 1/2 - 1/4 - 1/8...)
  8. What Question Must Be Asked?

    Just a small correction. No one can have 8 coins nor 2. If the total is 21: 5+7+9 is the only possibility. If the total is 4: 1+3+5 is the only possibility
  9. Soldiers in a field

    Cannot you shorten it? - if no one else watches A nor B, remove A and B and start over (this happens if A and B are very near and all other are far away enough) - if someone else watches A and/or B, at least one is not being watched (evident; if proof needed, start with 3 and continue with recursion)
  10. Who can go the lowest?

    Bad news guys, I win: But a mathematician will not agree.
  11. Protecting Airplanes

    What kind of planes are examined? And what kind of planes are not in the sample? https://en.wikipedia.org/wiki/Abraham_Wald
  12. Binary lock

    Real life: I bought (very cheep) a 256MB extension card for a 80286. There were 8 dip switches and no manual. I do not know how many times each switch can be flipped, in any case, it is better to minimize the manipulation. Look https://en.wikipedia.org/wiki/Gray_code
  13. In Soviet Russia

    Two judges meet in Moscow, one is laughing, laughing, laughing... The second asks why. 1st judge: I heard an awfully funny joke. 2nd judge: Tell me! 1st judge: I cannot. I sentenced the man who told it to 5 years.
  14. An Associative fallacy

    My favorite one: Take the sum of all the integers. Call it S. Take the sum of even integers. That sum is S/2. Because S is infinite, S/2=S, their difference is zero. So the sum of the odd integers is zero.
  15. Roll Them Out

    I googled some definitions of "middle" and "center". If this is THE solution, than the dictionaries are pretty wrong.
  16. Roll Them Out

    Lets try it together. Hidden Content Again, to obvious to be true. But what is wrong? It can be done in a smaller chute. Sure. Applying quantum physics.
  17. A loaded coin

    One possibility: However, in theory, it can go forever..
  18. Roll Them Out

    Lets try it together. Again, to obvious to be true. But what is wrong?
  19. Magical Tennis

    A typo again...I get used. Thinking it over, it is much more complicated.
  20. Magical Tennis

    So obvious that I fear I am missing something:
  21. You solved a particularly hard sudoku and you are eager to prove it. Just you do not want to give me any hint, nor, God bless, reveal the solution. How will you proceed?