Pebbling a Chessboard – Numberphile

Pebbling a Chessboard – Numberphile

100 thoughts on “Pebbling a Chessboard – Numberphile

  1. Why did we asume the original mass is 2? (its like starting with 1 pebble and associating it with the mass number 2, even though we started with 3 pebbles) If there were 3 clones originally, may i start the mass at 3? Or maybe even 6?

  2. It can't be done, even in an infinite # of moves, because the board still has a corner, and you can't put a clone in the corner.

  3. I tried to do this using a spreadsheet. You think it's simple enough, you get the first two out pretty easily, but the last one is desceptively tricky. You see that in order to get space for the clone, you need to duplicate this clone, but that is blocked by a third clone. Okay, so you duplicate the third one, then the second one, but then you realize that now the top part is blocked and you need to move that, so you go down the line, clone those, and you getmore and more blockages.

  4. The question was how to free the three prisoners from the prison.
    The answer was fill all the space outside the prison with prisoners.
    Now the prison is empty.

  5. And even if you had infinite turns, you'd still have to free the last clone… but there's no space… so you'd have to free up the area around it, which is just like freeing the original prison. So you'd have to free an infinite number of prisons, from the first to the infinitieth, in reverse order.
    And by that point, I suppose the entire board would be empty except for the one in the corner, since all the clones outside the prison would have to be in some "infinite" zone outside of the problem.

  6. This explanation is well done, but for those who are unsatisfied with some of the mysterious hand-waving and infinity arguments, don't worry. Using a similar idea, the proof can still be done by arguing for an upper bound of an n-by-n sized chessboard, and showing that for all n, the total area enclosed in the n-by-n chessboard is always less than 4.

  7. So, damn, amazing! I hate most of maths but when it's explained like this I'd love to know more. Extremely helpful.

  8. It feels like the proof mentioned in the video does not respect original rules of the game. The initial prison has 1,1/2 and 1/2. Now, one of the 1/2's splits into 2 quarters. But that disallows further splitting of the other 1/2, since for it to split, both its upper and right cells must be free! Hence, the 3rd row sum in the video cannot start with 1/4 :/

  9. I don't get the series thing. You're adding a positive value to a total an infinite number of times. So shouldn't you end up with infinity?

  10. Why do the "weights" of the clones have to reduce by half when cloned? That's critical to the proof but seems like an arbitrary assumption.

  11. I saw this question first time in Arthur Engel's Problem Solving Strategies (Chapter 1), the Q was slightly different though, very slight.

  12. If the rules say that no cloning allowed if the cell above or/and the cell to the right, then how is this [ 4:45 ] allowed?

  13. Numberphile, after seeing your, well, 2nd video I think, doesn't that prove that the original problem is unsolvable? There are only 2 clones on the first row or column after all, so the greenland is less than that of the prison! Are there any explanations to this?

  14. i constructed a scenario where just the two clones trying to make room for the one clone will fill an indefinitely large square leaving only their original spots and an indefinitely long strip are clear, so this should definitely be impossible

  15. if you can prove that (proving something impossible) is impossible, then that is a contradiction since you just proved something impossible

  16. I hav problems with the weighing methog. I would expect the square (1;1) to have the weight 1/2 since it gets weight 1/4 from two different squares. This would allow a much higher weight outside the prison than inside.

  17. If moving each clone takes a seconds (where a=the number assigned to the square the clone is on) won't the game finish in 4 seconds?

  18. Perhaps a simpler proof that you can't win would be this: the last move supposedly fills two holes but leaves one behind, but you can't fill that hole because you need two to make a move possible.

  19. Very elegant. Thank you very much. But now the new question: Can you free the clones in infinitely many moves? (With a one last move may be!) No! Because on the first row and the first column you can have at most two clones!

  20. A deity can solve the game. You just need sufficient 'power' over reality to do so. The game can be solved when time changes into eternity. The reality of eternity will instantaneously free the clones then and forevermore.

  21. Just wondering, but… is it possible to make it so that only one chip is left on the prison and that chip is on one of the "1/2" circle? Can't seem to do it :/

  22. Not to be rude or anything but I think I figured out that it is possible it would just take a really long time to do

  23. In reality, the sum does wind up equaling 2. The sum should equal 1.99999 repeating, which actually equals 2. So this problem should be solvable.

  24. A inverse geometric sequence is always convergent . The limiting function is always less than the whole. So they are not greater or equal to escape.

  25. It is also impossible to make sure that only one clone that is NOT the one at the bottom left remains in the prison. If you're truly interested in the proof, answer this comment

  26. It would still not be possible to get all the clones out, even with infinite moves.

    This is because when you make a move you cannot change the number of clones on the lowest row, or the leftmost column. So a maximum of two squares each on the lowest row and the left most column are occupied, with the others being empty.

    Therefore you have to leave a non-zero amount of space in greenland empty, so the space the clones can occupy in greenland is less than 2.

  27. It's also worth mentioning that the first column and row can never be completely filled, in fact they'll be empty except for one stuck in the second row on the first column and one in the second column on the first row (or just one clone in it's original position in the bottom left corner) and one clone at the farthest end you've reached in both the first column and row. You don't really need to go any further because as stated in the video as soon as one square remains empty you know the sum cannot add up to a whole number. A fraction will always be left behind in the prison no matter how many moves you make.

    If this still isn't clear: When starting you have three pieces all three HAVE to make a clone diagonally but only 2 can move vertically and 2 horizontally. Since none of the clones can jump down or to the right you cannot have more than 2 clones on the first row and column.
    Just wanted to highlight this since in the video it doesn't demonstrate this and leads you to believe that all squares could be filled in an infinite game which simply isn't possible.

  28. Explanation: we want to prove the game is impossible.
    Step 1) conspiracy theories on the total sum of the square
    Step 2) discovering the deep algorithm inside the game
    Step 3) say "the game is infinite, so it must be impossible"

  29. Heh, "pebbles". After "stones" from backgammon?
    Or, you could just call them "checkers", as we do in my country. 🙂

  30. The very first and crucial assumption a=2b seems completely arbitrary to me and a bit of a stretch. Surely this has been rigorously proven somewhere but that step did not make much sense to me, given how many different ways there are to fill up your board in practice.

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