Timeline for Reorienting a ladder among $\mathbb{Z}^2$ poles
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2017 at 11:16 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Broken link fixed.
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| Mar 21, 2017 at 15:45 | comment | added | Gerhard Paseman | Indeed, I am relating your metric to a pole bump metric. My point is that to handle a length 4 rod, I need to bump 4-2=2 poles, and each pole needs two moves to move the rod end around it. This bump metric may lead to a near optimal algorithm where the cost is number of translations and rotations. Gerhard "Excuse Me, I'm Coming Through" Paseman, 2017.03.21. | |
| Mar 21, 2017 at 11:47 | comment | added | Joseph O'Rourke | @GerhardPaseman: That may not be so different from the illustration, where I counted each translation as a separate move, e.g., position $10$ to $11$. Perhaps the metric should be: Each pure rotation about a fixed point is a move, each pure translation is a move. That's what I was using. | |
| Mar 21, 2017 at 10:58 | vote | accept | Joseph O'Rourke | ||
| Mar 21, 2017 at 1:25 | comment | added | Gerhard Paseman | For the example (starting from position 15), I rotate the rod clockwise around a lattice point and 1) bump into a pole, so 2,3)move the lower end around that pole with a translation and rotation, until I bump into the pole just above the first bumped pole, so I do 4,5) another translation and rotation to bring the rod to a 45 degree angle. This leads to a solution in 11 or fewer moves. Perhaps pole-bumping is an appropriate metric related to number of moves? Gerhard "These Poles Don't Blow Up" Paseman, 2017.03.20. | |
| Mar 21, 2017 at 0:49 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Mar 23, 2012 at 5:08 | comment | added | Joseph O'Rourke | @Gerry: $L=3.8$, $r=0.1$. Please note I do not know if 15 is the fewest number of "moves" to reorient. | |
| Mar 21, 2012 at 22:00 | comment | added | Gerry Myerson | What are the values of $L$ and $r$ for the shape in the 15-move diagram? | |
| Mar 21, 2012 at 19:13 | answer | added | Joseph O'Rourke | timeline score: 10 | |
| Mar 10, 2012 at 16:03 | comment | added | Joseph O'Rourke | @domotorp: Yes, that viewpoint is somehow clarifying. Thanks! | |
| Mar 10, 2012 at 9:01 | comment | added | domotorp | Of course the problem is equivalent to rotating a segment of length L with radius r obstacle-discs around each lattice point, for me its more natural to think about it this way. A natural upper bound on L is that it must fit in any angle - is this not sufficient? | |
| Mar 10, 2012 at 0:39 | comment | added | Joseph O'Rourke | Thank you, John, for recasting my question into such a perspicuous formulation! | |
| Mar 9, 2012 at 23:34 | comment | added | John Wiltshire-Gordon | Each legal position of the ladder can be described by the location of its center and its angle relative to the x-axis. Further, we do not mind translations by the integer lattice. This means that the configuration space of ladders is some closed semi-algebraic subset of the torus T^3. We are trying to understand the connected components. | |
| Mar 9, 2012 at 21:34 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added a second illustration.
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| Mar 9, 2012 at 14:24 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |