Timeline for Can the fugitive escape?
Current License: CC BY-SA 4.0
28 events
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| May 26, 2021 at 19:05 | comment | added | jcdornano | Indeed, it is the case. It can ne deduced from the proof I gave un my answer. As well as the case $c<2$. | |
| Apr 23, 2021 at 15:42 | comment | added | jcdornano | I even think that the the fact that one officer is aloud to move exactly delta is better for them than any $c$-game for $c$ as big as we want , but with the condition that each officer moves canot be more than $\delta-\epsilon $ for any $\epsilon>0$ ... | |
| Apr 23, 2021 at 15:34 | comment | added | jcdornano | I think this strattegy is not working anymore if c=2 but officers move are strictly less than $\delta$. Because if F is not moving, and the officer are shrinking the net, then F will décide to move if and only officer are at a distance that is at most $\delta$, as soon at the distance is less , F jumps out, and the cortical case is when the distance is exactly $\delta$, then he will be caught for sure if and only if one officer is cloud to move $\delta$. | |
| Apr 23, 2021 at 14:54 | comment | added | jcdornano | You are right^^, but then they have to shrink it wisely... if they shrink it to much, F can escape in one move out of the triangle | |
| Apr 23, 2021 at 14:09 | comment | added | Pace Nielsen | @jcdornano If the fugitive doesn't move, the officers use all their speed to shrink the net. | |
| Apr 23, 2021 at 13:59 | comment | added | jcdornano | Same if the fugitive moves are $\epsilon^{-n}$ to move $n$ for a small enough fixed $\epsilon$ | |
| Apr 23, 2021 at 13:13 | comment | added | jcdornano | Pace Nielsen, strictly speaking, the strattegy you posted in your answer is winning for the fugitive ... if he never moves... | |
| Apr 23, 2021 at 11:17 | comment | added | Eric | @usul Here's an answer (a long one, scroll to the end) that proves if $c\lt 1$ then the fugitive escapes: mathoverflow.net/a/328213/75935 | |
| Apr 21, 2021 at 2:10 | history | edited | Eric | CC BY-SA 4.0 |
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| Apr 21, 2021 at 1:58 | history | edited | Eric | CC BY-SA 4.0 |
adding the direction to explore c from below
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| Apr 20, 2021 at 19:21 | comment | added | usul | @Eric For a lower bound, I'd start by assuming the officers are equally spaced on a perfect circle. Generic configurations sound very difficult! | |
| Apr 20, 2021 at 16:30 | history | made wiki | Post Made Community Wiki by Stefan Kohl♦ | ||
| Apr 20, 2021 at 16:28 | comment | added | Eric | I think you're right. I forgot that they're not using their full movement before reaching the corner. Maybe the other direction is also worth exploring: the lower bound of $c$, i.e., finding some $c\lt 1$ such that the fugitive can escape. | |
| Apr 20, 2021 at 16:03 | comment | added | Pace Nielsen | @WillSawin Regarding your question: Suppose the fugitive only moves parallel to one edge. The officer on that parallel edge has to use his full movement speed just to maintain his (extremely small) lag. The other two officers have a bit more freedom, especially the one opposite his movement, but let's ignore that and suppose they also use the rest of the speed just to maintain their lag. When the fugitive gets sufficiently close to an edge the officer on that edge can just deviate and catch him. | |
| Apr 20, 2021 at 15:52 | comment | added | Pace Nielsen | ...catch up. Any movement off diagonal gives the officers extra movement speed. So if the lag of each officer is $\epsilon$, they can choose this $\epsilon$ (depending on the box size, and depending which diagonal the fugitive chooses to follow) small enough so that the fugitive cannot leave the diagonal far enough to get away from the corner sufficiently far to avoid capture. | |
| Apr 20, 2021 at 15:47 | comment | added | Pace Nielsen | @Eric This is somewhat moot now that Will has a better constant, but it isn't as hard as you make it. First, let's agree that once they lag a bit, they no longer ever shrink the box, but work to only catch up. Second, I'll leave it to you to show that if the fugitive gets too close to one of the corners, then the two officers close-by can use all the movement speed and catch the fugitive. So, the only question is whether or not the fugitive can veer off course and avoid the corner, while also not giving the officers enough extra movement speed (because he didn't go diagonal) for them to... | |
| Apr 20, 2021 at 13:21 | comment | added | Will Sawin | @PaceNielsen I don't see how to do $c=2$ here yet. The directions where the bound is sharp are those parallel to an edge. As soon as any of the officers lags a little, if the fugitive keeps moving in these directions, the officers have to use their full movement to follow and so will not be able to reduce the amount of lag. | |
| Apr 20, 2021 at 13:13 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Apr 20, 2021 at 12:20 | comment | added | Eric | @usul Yes, the boundary cases are bound to be tricky, and lagging dicey. In fact, $2 \sqrt2$ won't work even if you place four additional officers to guard the corners. | |
| Apr 20, 2021 at 12:10 | comment | added | usul | @Eric this makes sense. To clarify, you agree with $c > 2\sqrt{2}$, but are questioning $c=2\sqrt{2}$? | |
| Apr 20, 2021 at 11:53 | comment | added | Eric | Without the lag they can surely use that extra movement to shrink the box. But now they have to shrink and close the lag in time. I do think the situation is very subtle here. But even without shrinking, lagging can be problematic. For example, if $2\delta$ divides the diagonal of the starting rectangle of the officers, then any lag will prevent the officers to capture the fugitive at the diagonal corner. | |
| Apr 20, 2021 at 5:51 | comment | added | Pace Nielsen | @WillSawin Go ahead and turn my answer into a community answer, and update it with your solution. (I bet you could get c=2 as well, throwing in a slight lagging argument.) | |
| Apr 20, 2021 at 5:48 | comment | added | Pace Nielsen | @Eric The more he goes off diagonal, then more extra movement speed they get to catch up and box him in. The larger the box, the smaller the lag, so tit only takes a tiny amount of deviation from the diagonal to catch back up. | |
| Apr 20, 2021 at 3:15 | comment | added | Eric | If officers lag a little behind, the fugitive can deviate from the diagonal. What are officers to do then? Are you sure they can still squeeze as much as they want over a finite period of time? | |
| Apr 20, 2021 at 2:52 | comment | added | Will Sawin | @usul I'm pretty sure 3 officers on the 3 sides of an equilateral triangle, each one staying on the orthogonal projection of the fugitive to their side, can handle any $c>2$, since $\max ( | \cos (\theta) | + |\cos(\theta+ 2\pi/3)| + |\cos(\theta+ 4\pi/3)|| ) =2$. | |
| Apr 19, 2021 at 17:00 | comment | added | Pace Nielsen | @usul No idea how to improve it. The idea came from thinking about the rook version. Perhaps someone who plays hexagonal games will have another idea. | |
| Apr 19, 2021 at 16:34 | comment | added | usul | Very nice! Any thoughts on modifying this scheme with 6 officers / a hexagon? | |
| Apr 19, 2021 at 15:09 | history | answered | Pace Nielsen | CC BY-SA 4.0 |