Timeline for Conjecture regarding closest point inside a discrete ball to a line
Current License: CC BY-SA 3.0
23 events
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| Nov 23, 2014 at 9:27 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 23, 2014 at 9:24 | comment | added | Hao Chen | @Rob. I'm fully aware of its twin. What I missed in domotrop's proof is that this twin must be in the first quadrant, then the proof works by replacing $y$ by $x$ when constructing the parallelogram. | |
| Nov 23, 2014 at 9:19 | comment | added | Hao Chen | @domotrop. Thank you for explanation. So I imagine that you repeat the same argument to the part of the first quadrant above $\ell$, but replace $y$ by $x$ when constructing the parallelogram. This then covers all the angles. That's also why you can suppose $P$ is below $\ell$ wlog. Then I think the detail that bothers me is "$P$ minimizes the angle", which is actually the "angle" from below. I'm now totally OK with the proof. | |
| Nov 23, 2014 at 9:12 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 23, 2014 at 6:03 | comment | added | domotorp | The last sentence of my proof is supposed to cover such cases, as that of the blue point. Any integer point in another quadrant has distance $>1$ from the line, so larger than, say, $(1,1)$. | |
| Nov 23, 2014 at 1:38 | comment | added | Rob | If such a point exists, it has to have a twin in the upper half plane, where we know the proof works. The difference in angle between the point and its twin is $\pi$. | |
| Nov 23, 2014 at 1:31 | comment | added | Hao Chen | @Rob, my concern is, in your argument, there might be a closer point in lower half space, whose angle is not any of the four candidates. | |
| Nov 23, 2014 at 1:29 | comment | added | Hao Chen | @Rob, indeed, but that's not my concern, I understand what you mean. | |
| Nov 23, 2014 at 1:26 | comment | added | Rob | You are right - the correct statement should not be that the minimizer $n+im$ obeys $Arg(n+im) \in \{\theta_i,\theta_{i+1}\}$, but rather $Arg(n+im) \in \{\theta_i,\theta_{i+1},\theta_i+\pi,\theta_{i+1}+\pi\}$. But perhaps a nicer way of saying things is that the above result gives you the solution with $Arg(n+im) \in [0,\pi)$, from which a second solution $-n-im$ in $[\pi,2\pi)$ is constructed. | |
| Nov 23, 2014 at 1:13 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 23, 2014 at 0:51 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 23, 2014 at 0:15 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 23:47 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 23:41 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 23:34 | history | undeleted | Hao Chen | ||
| Nov 22, 2014 at 23:34 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 22:51 | history | deleted | Hao Chen | via Vote | |
| Nov 22, 2014 at 22:47 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 22:47 | history | undeleted | Hao Chen | ||
| Nov 22, 2014 at 22:46 | history | deleted | Hao Chen | via Vote | |
| Nov 22, 2014 at 22:34 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 22:29 | history | edited | Hao Chen | CC BY-SA 3.0 |
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| Nov 22, 2014 at 22:23 | history | answered | Hao Chen | CC BY-SA 3.0 |