Timeline for Polygonal paths and polygons with prescribed set of vertices
Current License: CC BY-SA 4.0
33 events
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| Dec 7, 2019 at 22:04 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 21:58 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 20:32 | answer | added | Dmitri Panov | timeline score: 5 | |
| Dec 7, 2019 at 20:27 | history | edited | Per Alexandersson | CC BY-SA 4.0 |
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| Dec 7, 2019 at 20:24 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 16:54 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 16:48 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 16:09 | comment | added | Gerhard Paseman | For n=1, Joseph has a diagram of impossibility. Likely a more complicated one exists for n=2. However, there is a path for n=3 and thus for all higher n. Break into the union of a central point and 3 of (n=3/2) size triangles with overlap. One has a path from the center filling the triangles in a cyclic order, and can end on an external vertex. Gerhard "Seeing Spots Before My Eyes" Paseman, 2019.12.07. | |
| Dec 7, 2019 at 14:59 | comment | added | Algirdas Rugys | @M. Winter. In the given example, A for fixed n∈N, consists of (n+1)(2n+1) points with nonnegative integer coordinates. But you can't move them arbitrarily. Look, please, at the pictures of J.O'Rourke, below. | |
| Dec 7, 2019 at 12:09 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 11:27 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 7, 2019 at 2:16 | history | edited | user44143 | CC BY-SA 4.0 |
simplified and standardized the phrasing
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| Dec 7, 2019 at 2:02 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| Dec 7, 2019 at 1:14 | comment | added | Joseph O'Rourke | Is "a broken line" what would normally be called a (closed) simple polygon (an 𝑛-gon as per @Dmitri), or is it instead an open, simple polygonal path? The term "broken line" does not have a standard definition in the (English) literature. | |
| Dec 7, 2019 at 1:07 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 6, 2019 at 23:31 | comment | added | Gerhard Paseman | I conjecture that there are only finitely many even numbers for which the given configuration does not have a path. Further, one can do a breadth first search to find such a path. Since you ask for no intersecting lines, you can reduce consider a divide and conquer algorithm by choosing successive vertices as midpoints of the end path. I don't know how fast it will be. Gerhard "Computer Can Handle Size Six" Paseman, 2019.12.06. | |
| Dec 6, 2019 at 16:37 | answer | added | Joseph O'Rourke | timeline score: 4 | |
| Dec 6, 2019 at 16:20 | comment | added | Algirdas Rugys | I conject, that if we take any even number instead of 4 in the example, the answer remains negative. It's interesting to which even numbers computer could check. | |
| Dec 6, 2019 at 15:12 | comment | added | user44143 | I agree that your example of the fifteen points does not have a broken line, but I don’t see a clean proof. | |
| Dec 6, 2019 at 13:33 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 6, 2019 at 13:14 | comment | added | Algirdas Rugys | P.S. $x , y$ are nonnegative integers | |
| Dec 6, 2019 at 13:06 | comment | added | Algirdas Rugys | Take A 15 points (x;y) with $0\leq x+y \leq 4$ . It seems, that there isn't such a broken line. But how to prove this ? | |
| Dec 6, 2019 at 12:33 | comment | added | Dmitri Panov | Fedor, since the modified question is now asking the angles not to be straight, this is not a duplicate. This version if obviously harder. | |
| Dec 6, 2019 at 11:57 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 6, 2019 at 11:56 | history | undeleted | Algirdas Rugys | ||
| Dec 6, 2019 at 11:55 | history | deleted | Algirdas Rugys | via Vote | |
| Dec 6, 2019 at 11:52 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 6, 2019 at 11:43 | comment | added | Fedor Petrov | duplicate mathoverflow.net/q/226469/4312 | |
| Dec 6, 2019 at 11:34 | history | edited | Algirdas Rugys | CC BY-SA 4.0 |
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| Dec 6, 2019 at 11:13 | comment | added | Dmitri Panov | @Gerry, I see. If such a point doesn't count, the answer will be considerably more involved indeed (I don't know the answer for such a version of the question). I guess, Algirdas should add this information to the question if he wants to exclude this type of vertices. | |
| Dec 6, 2019 at 10:58 | comment | added | Gerry Myerson | @Dmitri I don't know whether a point counts as a vertex if the angle there is a straight angle. | |
| Dec 6, 2019 at 10:11 | comment | added | Dmitri Panov | Such an $n$-gon exists always provided the points don't lie on one line. Suppose that there are $3$ points $p, q, r\in A$ that don't lie on one line. Take a generic point $O$ inside the triangle $pqr$, such that no line containing two points from $A$ passes through $O$. Once you do this, you can enumerate all the points of $A$ in the anti-clockwise order with respect to $O$, $A=p_1,\ldots, p_n$. Then you join each $p_i$ with $p_{i+1}$ by a segment (and $p_n$ with $p_1$). This will clearly give you the desired $n$-gon. | |
| Dec 6, 2019 at 9:23 | history | asked | Algirdas Rugys | CC BY-SA 4.0 |