Timeline for Polygonal paths and polygons with prescribed set of vertices
Current License: CC BY-SA 4.0
15 events
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| Dec 8, 2019 at 2:05 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Dec 7, 2019 at 22:21 | comment | added | user44143 | You could add: This path can be extended with a sawtooth pattern on each pair of rows below it, showing that there are appropriate paths on any larger triangles with even side lengths. | |
| Dec 7, 2019 at 21:35 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Dec 7, 2019 at 21:16 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Dec 7, 2019 at 17:26 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Dec 7, 2019 at 2:13 | comment | added | Gerhard Paseman | By affine invariance , the poster's suggested configuration is equivalent to the problem on a triangular grid. There one has that an extreme vertex must not be an endpoint, and with k=2 one has two cases to check, both of which isolate a vertex under the conditions. Gerhard "Let Symmetry Do The Work" Paseman, 2019.12.06. | |
| Dec 7, 2019 at 1:28 | comment | added | Joseph O'Rourke | @GerhardPaseman: The OP clarified---Your interpretation is what he intended, mine was incorrect. Still my example shows there is no simple polygonal path through those $6$ points, turning at each. | |
| Dec 7, 2019 at 0:43 | comment | added | Joseph O'Rourke | @GerhardPaseman: Ah, and I was assuming a "broken line" is a polygon. Your interpretation may be the correct one. Clearly there is confusion over this issue. | |
| Dec 7, 2019 at 0:40 | comment | added | Gerhard Paseman | I interpret a broken line as a broken line segment, which has the (topological) shape of a path, not of a cycle. Gerhard "Who Sometimes Does Not Return" Paseman, 2019.12.06. | |
| Dec 7, 2019 at 0:35 | comment | added | Joseph O'Rourke | @GerhardPaseman: What does it mean for a path to end at a vertex, when the path must be a closed cycle? | |
| Dec 6, 2019 at 23:17 | comment | added | Gerhard Paseman | You can reduce this to whether a path ends at vertex 1 or not. When it doesn't, symmetry gives two cases that are easily checked. A slow divide and conquer algorithm can handle the general problem in a similar way. Gerhard "Knows No Fast Algorithm Yet" Paseman, 2019.12.06. | |
| Dec 6, 2019 at 16:50 | history | undeleted | Joseph O'Rourke | ||
| Dec 6, 2019 at 16:50 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Dec 6, 2019 at 16:45 | history | deleted | Joseph O'Rourke | via Vote | |
| Dec 6, 2019 at 16:37 | history | answered | Joseph O'Rourke | CC BY-SA 4.0 |