Jump to content
BrainDen.com - Brain Teasers


  • Posts

  • Joined

  • Last visited

  • Days Won


Everything posted by harey

  1. Congratulations, works. I checked it: I suspect a kind of recursivity, but it probably will not show up for small n. For n=2 and n=3, the robot stays where it should, which does not seem evident for larger n.
  2. I do not really understand what you mean by "simple path", but I think the answer is no. The figure H is OK, but you can very well have: * * * * * c * * * c * c c * * * c * * * * * * * * c * ... Your list must work for all these cases, as well as for all grids where there is/are one/three/four cement blocks. i.e. the solution Molly Mae proposed would not work for the first maze (so it does not matter anymore that it works for the 2nd and 3rd). Even calculating the number of possibilities for 2 blocks gives me headaches. 7 * 6 / 2? Wrong, the robot must be free to leave the corner and the lower right corner must remain accessible. To get insane...
  3. An interesting variant. The problem states "there is a path", but it does not state "there is a path for the robot". I think it is easier we stay with moving just to the next square.
  4. Each square of an n x n grid of squares is either filled with cement blocks or left empty, such that there is at least one path from the top left corner to the bottom right corner of the grid. Outside the grid everything is filled with cement. A robot is currently located at the top left corner and wants to get to the bottom right corner, but it only knows the value of n and doesn't know the layout of the grid. It also has no method of observing its surroundings, and it is your job to give it instructions to ensure it ends up at its destination. Your instructions should be a finite list of directions (Up, Down, Left, Right) - the robot will try to move in the indicated directions in order, and, if there is a cement wall in the way at some step, it will simply fail to move in the corresponding direction and continue on with the next instruction in the list. Since the robot has no way of sensing whether it has reached its destination, it might reach the destination somewhere in the middle of your list of instructions and then later leave. The goal is to give a list of instructions, depending only on n, such that after following your instructions the robot is guaranteed to end its journey in the bottom right corner of the grid. The bad news: I do not know the solution and I cannot ask for hints.
  5. Evident after translating it to college version: Just a little bit late... I guess in a test, half an hour would be allocated for this question. I get depressed when I realize the time I needed.
  6. The plot for all cases: Strategy I knew it would be hard, but not THAT hard.
  7. There are 9 cards on the table, numbered 1 to 9, face up. You and I alternatively pick one card. The first one who can produce exactly 3 cards summing up exactly 15 wins. What is your strategy?
  8. I found a strategy, but far from sure it is optimal.
  9. True so far. But it does not imply that it is not advantageous to i.e. wait for the 3 next cards and call "Stop!" After the next 3 cards, the p(next card is R)=p(the last card is R) still holds - but with (most probably) a different p.
  10. I do not think there is need to use the spoiler anymore. If the blue are excluded because every room houses a green man, the same way you cannot accept even a single new guest.
  11. 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.
  12. 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?
  13. 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?
  14. 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?
  15. 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?
  16. 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
  17. 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...)
  18. 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
  • Create New...