Timeline for Can an odd number of marbles jump to infinity?
Current License: CC BY-SA 4.0
19 events
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May 4, 2022 at 16:27 | history | edited | Saúl RM | CC BY-SA 4.0 |
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May 4, 2022 at 14:17 | comment | added | Cœur | the only number... is 3... or 1 | |
May 3, 2022 at 9:59 | comment | added | domotorp | Makes perfect sense, in fact, I see that we can similarly attach two marbles to a 3 by 3 square, so mod 3 doesn't matter. I suppose the last thing to prove is that less than $q^2$ marbles are not enough if one is allowed to move $q$ at a time. | |
May 3, 2022 at 8:59 | comment | added | Saúl RM | @domotorp it seems there is an analogous construction using $10$ pieces, a marble attached to a $3\times3$ square of marbles (instead of a $2\times 2$ one as in the answer) | |
May 3, 2022 at 8:43 | comment | added | Saúl RM | @FrançoisBrunault true, and $(0,2)$ is achievable if we do not restrict to up/right moves (see the answer before the edit). So $(2,0)$ can be obtained too conjugating by a transformation of ``` ... .oo ooo ``` into ``` .o. .oo .oo ``` | |
May 3, 2022 at 7:55 | comment | added | François Brunault | $(1,0)$ is not possible, by a parity argument: the sum of the $x$-coordinates over all marbles keeps the same parity, and there is an odd number of marbles. So the question left is whether $(2,0)$ is achievable. | |
May 3, 2022 at 7:37 | comment | added | François Brunault | One can also ask whether the group of marbles can reach any location in the infinite chess board (and more precisely, which translations are achievable from the initial position). In your solution, you have translation by $(2,2)$. If I'm correct it's easy to adpat your idea to get translations by $(\pm 2, \pm 2)$ (independent signs) but I don't know if e.g. $(1,0)$ is possible. | |
May 3, 2022 at 4:35 | comment | added | domotorp | By 'moving 3 marbles at a time' I mean that a marble is allowed to jump over exactly two other marbles that are next to it in the same row/column. | |
May 3, 2022 at 2:57 | comment | added | domotorp | Amazing! Interesting that the period is this long. I assume that the next natural question would be to find a set of marbles not divisible by 3 that can go to infinity, if we are allowed to move 3 marbles at a time. Would some simple modification of your construction work? | |
May 2, 2022 at 23:27 | comment | added | Saúl RM | I assumed that with "jumping to infinity" he meant that all the marbles eventually leave any bounded set | |
May 2, 2022 at 23:25 | comment | added | Timothy Chow | Is there a tacit condition that the set of marbles is supposed to retain the same "shape" as it goes to infinity? Otherwise, if going off to infinity is the only condition, then to any finite configuration, we can just add 2 marbles somewhere that march off to infinity independently of what the other marbles are doing. | |
May 2, 2022 at 23:18 | comment | added | Saúl RM | Actually I just thought of this counterexample and that it implies that given $n\geq4$ there is a set of $n$ marbles which can go to infinity. The second question seems interesting even if you only allow moves up/to the right | |
May 2, 2022 at 22:59 | comment | added | Joseph O'Rourke | Could you sketch an argument that shows that configurations of larger size can behave the same way? And might there also be disjoint configurations that unavoidably interfere with one another? | |
May 2, 2022 at 22:40 | comment | added | Yuval Peres | Wow! beautiful. | |
May 2, 2022 at 20:37 | comment | added | Saúl RM | I see, I included a new sequence which only has those moves (although it is a bit more complicated) | |
May 2, 2022 at 20:34 | history | edited | Saúl RM | CC BY-SA 4.0 |
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May 2, 2022 at 20:09 | vote | accept | domotorp | ||
May 2, 2022 at 20:05 | comment | added | domotorp | That's a nice example! To be fair to myself, I had an extra condition that the allowed jumps are only upwards and to the right (see motivation), but when posing the question, I didn't include this. | |
May 2, 2022 at 19:37 | history | answered | Saúl RM | CC BY-SA 4.0 |