Timeline for The most number of points that realize only $k$ distinct distances
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 10 at 0:27 | comment | added | Oscar Lanzi | We certainly have $f_3(3)\ge12$ with the regular icosahedron. | |
| May 25, 2015 at 7:18 | comment | added | Aaron Meyerowitz | Of course I was wrong. | |
| May 24, 2015 at 17:13 | vote | accept | Joseph O'Rourke | ||
| May 24, 2015 at 17:03 | history | edited | GH from MO | CC BY-SA 3.0 |
small TeX fix
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| May 24, 2015 at 16:26 | answer | added | Anurag | timeline score: 9 | |
| May 24, 2015 at 16:09 | answer | added | Kristal Cantwell | timeline score: 3 | |
| May 24, 2015 at 15:53 | comment | added | Joseph O'Rourke | @SamHopkins: You are right, I need to study the Guth-Katz bound. | |
| May 24, 2015 at 15:50 | comment | added | Sam Hopkins | @JosephO'Rourke: Have you looked into the many recent techniques and approaches developed to attack the asymptotics of this problem? For instance it is still probably worthwhile to convert this to an incidence question. | |
| May 24, 2015 at 15:44 | comment | added | The Masked Avenger | If Aaron is right, that would make the face angle of a regular tetrahedron exactly 72 degrees. Which it isn't. | |
| May 24, 2015 at 13:44 | comment | added | Joseph O'Rourke | @AaronMeyerowitz: Hmm. For a pentagon inscribed in a unit-radius circle, I find $s \approx 1.176$ while the distance from each added above/below point to a pentagon point is $s' \approx 1.160$. | |
| May 24, 2015 at 7:22 | comment | added | Aaron Meyerowitz | @Anthony is right. It so happens that the two extra points are distance $s$ from each other and from each of the pentagon points where the pentagon has side $s$. | |
| May 24, 2015 at 4:51 | comment | added | Anthony Quas | Doesn't the pentagon give you $f_3(2)\ge 7$ if you add a point below also? | |
| May 24, 2015 at 3:46 | comment | added | Will Brian | Also, $f_d(k) < R(d+2,\dots,d+2)$ (where $R$ denotes the Ramsey number and there are $k$ entries). This is because you cannot have $d+2$ points that are all mutually the same distance from each other. Suppose you had $R(d+2,\dots,d+2)$ or more points and only $k$ distances represented. Think of these points as the vertices of a complete graph, and think of the distance between two points as the "color" of their edge. The definition of $R$ tells you that you have $d+2$ points all the same distance apart, a contradiction. Thus, for example, $f_3(2)$ is less than $R(5,5) \leq 49$. | |
| May 24, 2015 at 3:23 | comment | added | Will Brian | Don't know if this helps, but the octahedron is another example showing $f_3(2) \geq 6$. | |
| May 24, 2015 at 2:01 | comment | added | Joseph O'Rourke | @SamHopkins: Yes, I think Gerhard's comment clarifies. The Erdős conjecture is that the min # of distances is $g_d(n)=\Theta(2^{n/d})$, which reverses to saying that my $f_d(k)$ grows like $d \log k$. This is already useful (Thanks!) but doesn't help me determine, e.g., $f_3(2)$... | |
| May 24, 2015 at 1:54 | comment | added | Gerhard Paseman | It is related, but not exactly the same. In Joseph's version, k (number of distinct distances) is fixed and n is wanted, whereas the literature you mention seems to me to have n fixed and estimates k given n and d. However, that entry a good place to start. Also, the book titled something like "Unsolved problems in geometry" might have a discrete portion that gets closer to Joseph's question. Gerhard "Looking From The Other End" Paseman, 2015.05.23 | |
| May 24, 2015 at 1:41 | comment | added | Sam Hopkins | Is this not a very well-known problem?: en.wikipedia.org/wiki/Erd%C5%91s_distinct_distances_problem | |
| May 24, 2015 at 1:36 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |