Timeline for VC dimension of full-dimensional closed polyhedral cone in $\mathbb R^d$
Current License: CC BY-SA 4.0
12 events
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| Apr 16 at 6:32 | comment | added | domotorp | I'm again getting a bit lost with the definitions. I think it would be best if someone else answered your question. | |
| Apr 15 at 16:36 | comment | added | Neophyte | In general, if we take a set of $d$ independent vectors and apply an affine transformation to transform them into $e_1,\cdots, e_d$, it need not be the case that the hypothesis that realize the labeling vector $l \in \{-1,1\}^d$ lie in $C$ itself. For $\mathbb R^d$ cone, this was trivial but need not be the same for a non-trivial full dimensional polyhedral cones like pointed ones. | |
| Apr 15 at 16:30 | comment | added | Neophyte | I think the issue is that the cone $C$ can be restricted/pointed and need not contain all hypotheses that would shatter the d independent points. For example, in 2D, we can clearly see that no two points can be shattered by a pointed full dimensional cone - example. So, the VC of that 2D cone seems to align with $VC = d-1 = 1$. | |
| Apr 13 at 4:42 | comment | added | domotorp | I agree that the cited answer by @domotorp is quite hard to read. But why can't you take an affine transformation that converts any independent $x_1\ldots,x_d$ to $e_1,\ldots,e_d$ to get an orthant? Now it seems to me that the VC-dim will be even $d$, I don't get why the other answer claims only $d-1$. | |
| Apr 10 at 20:04 | history | edited | Neophyte | CC BY-SA 4.0 |
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| Apr 10 at 20:03 | history | edited | Neophyte | CC BY-SA 4.0 |
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| Apr 10 at 19:28 | history | edited | Neophyte | CC BY-SA 4.0 |
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| S Apr 10 at 19:16 | review | First questions | |||
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| S Apr 10 at 19:16 | history | asked | Neophyte | CC BY-SA 4.0 |