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bonanova

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Everything posted by bonanova

  1. @CaptainEd Wow. Not the answer I had in mind -- a much better one! Nice. Bonus (slightly modified) version. Same question, but this time no cut may end short of the opposite edge. It must go entirely through the piece being cut.
  2. @harey Hilbert's hotel tells us not to treat countably infinite sets the same way we treat finite sets. There is no 1-1 correspondence between { 1 3 5 7 9 } and { 1 2 3 4 5 6 7 8 9 10 } as there is between the (infinite set of) { odd integers } and the (infinite set of) { integers.} Hilbert's hotel always has room for more. Completion (of an infinite set of tasks) is another tricky concept. If we number some tasks 1 2 3 ... and there is no final integer, how can there be a final task? And if there's no final task how can we complete infinitely many tasks, or describe the state of things after they have transpired? We finesse that point with a 1-1 correspondence of event times to the terms of an infinite series, 1 1/2 1/4 1/8 ... 1/2^n ...., which (conveniently) converges. We pack a countably infinite set of events into a time interval that ends at midnight. "Completion" is cleverly accomplished in two seconds. It's non-physical enough that it never could happen, but we can reasonably discuss the post-midnight state of affairs nonetheless. We have two countably infinite sets, {coins} and {events}. At each event, two coins are added to a box and one coin from that box is discarded. So, of the two added coins, at least one is kept. We know that at midnight the entire set {coins} has entered each box, and some (proper or not) subset of them has been discarded. The key question is this: can a coin that is kept at a certain event ever be discarded at a later event? For Al, the answer is yes. Al always discards his lowest coin, so at each event time ti he discards coin ci. Thus eventually that is, upon completion of the infinite set of tasks, every coin that is initially kept is later discarded. At midnight no coins remain. Al's box is empty. For Bert the answer is no. Bert discards the highest-numbered of his coins, and that is always the even coin that he just received. At no event is Bert ever scheduled to discard an odd coin. Every odd coin that enters Bert's box is kept, and it stays there forever. Bert's box contains a countably infinite set of coins. For Charlie the answer is ... well ... um ... actually ... I guess ... yeah, but it might take an infinite number of events for it to be discarded. Well it just so happens that we have an infinite number of events that follow the keeping of every one of Charlie's initially kept coins. So, yes. All of Charlie's coins that are not immediately discarded are eventually discarded. At midnight, Charlie's box is empty.
  3. That would not be permitted. The OP says with each use of the saw you may pick up one piece of the board and make a straight cut.
  4. You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
  5. Gah! Mine was not a useful clue. The denominator (greater than 6 but not huge) is too large for a simulation to tell you the fraction. Also I think your simulation value is high. But you're thinking in the right direction. Here may be more useful clues.
  6. Four pegs begin at the corners of a unit square on a grid having integer coordinates. At any time one peg may jump a second peg along any straight line and land an equal distance on its other side. The jumped peg remains in place. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + O + + + + + + O O + ==> + + + O O + + + + O O + + + + O + + + + + + + + + + + + + + Is it possible to maneuver the pegs to the corners of a larger square? + + + + + + + O + O + + + + + + + + + O + O + + + + + + + + + + + + + +
  7. Nice. A construction is certainly a proof of existence. A pairing without intersects exists and ... here it is! Now I wonder if for any groups of n blue and n red points there is only one pairing without crossings?
  8. One approach is to Then it would be helpful to know that the result is
  9. plainglazed suggests an analysis amenable to powers of 2. I'll amend the OP to make it 1024 cars instead of 100 cars.
  10. Actually, two of the boxes will be empty.
  11. @Pickett Interesting result. OP meant for the coins to have equal value, but yours is a great puzzle also, even for finite numbers of coins received. @rocdocmac and @Pickett When midnight has struck an infinite number of events will have transpired and the process will have stopped. Will any of the three be happier than the others?
  12. At an ever increasing pace Al, Bert and Charlie have been receiving into three identical boxes of limitless capacity identical pairs of silver coins engraved with the integers 1, 2, 3, 4, ... etc. These events occur on a precise schedule, each box receiving Coins marked 1 and 2 at 1 minute before midnight Coins marked 3 and 4 at 1/2 minute before midnight Coins marked 5 and 6 at 1/4 minute before midnight Coins marked 7 and 8 at 1/8 minute before midnight etc. But they were instructed at each event to remove a coin from their respective boxes and discard it. After some thought, Al decided each time to discard his lowest-numbered coin; Bert discarded an even-numbered coin; and Charlie thought what the heck and discarded a coin selected at random. Regardless of strategy, at each event the number of coins in each box grew by unity, so that after N events each box held N coins. Needless to say when midnight struck their arms were infinitely tired, but it was a small price to pay for infinite riches. But tell us, now, whether their expectations were met. Describe the contents of each box at midnight.
  13. Not bad. I like the fact you can throw away the connections you find without fear of removing a potential crossing with a future connection. It’s not the solution I had in mind but I think it works. Nice.
  14. Just an observation, that may already have been uncovered, but don't know where to go with it:
  15. Final clue Repeating an earlier comment,
  16. Belt "segment" just refers to portions of belt between adjacent bags. Top or bottom is not an issue. Consider bags are randomly placed points on a line if you like. The fraction of length that is not a near-neighbor segment changes with time. To imply a limit, the OP asks "on average" over time.
  17. OK a clue ( and it was plainglazed third answer that's close )
  18. plainglazed is closest so far (but I can’t say which of his guesses). Remember it’s a physics question.
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