Timeline for Representing a binary relation
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2022 at 2:51 | vote | accept | Arthur B | ||
| Nov 11, 2022 at 13:03 | answer | added | Emil Jeřábek | timeline score: 2 | |
| Nov 4, 2022 at 22:03 | comment | added | Arthur B | @EmilJeřábek I'm not sure how you would construct $g$, can you elaborate? | |
| Nov 4, 2022 at 22:02 | comment | added | Arthur B | @GerryMyerson it's a popular term in machine learning, en.wikipedia.org/wiki/One-hot has a few references. | |
| Nov 3, 2022 at 9:32 | comment | added | Emil Jeřábek | I haven’t tried to flesh this out, but what if I take $d=2$ and $g(x_i)=(i,-i)$, where $X=\{x_i:1\le i\le n\}$? Since the $g(x_i)$ are pairwise incomparable, I think it should be possible to construct continuous $f\colon\mathbb R^4\to\mathbb R$ nondecreasing in all coordinates that takes arbitrary given values on the pairs $(g(x_i),g(x_j))$ (so I don’t even need any assumptions on the relation $R$). Or is there something seriously wrong with this idea? | |
| Nov 3, 2022 at 7:50 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 2, 2022 at 12:22 | comment | added | Joel David Hamkins | Sorry, should be spelled asymmetric. | |
| Nov 2, 2022 at 11:54 | comment | added | Joel David Hamkins | Relation $R$ is antisymmetric means: $xRy\wedge yRx\implies x=y$. Relation $R$ is assymmetric means: $xRy\implies \neg yRx$. | |
| Nov 2, 2022 at 11:27 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Nov 2, 2022 at 11:26 | comment | added | Joel David Hamkins | I think you mean asymmetric+irreflexive, rather than anti-symmetric+nonreflexive. | |
| Nov 2, 2022 at 8:48 | comment | added | Gerry Myerson | So, one-hot encoding means "a one-one map from a set of $n$ elements to the canonical basis of ${\bf R}^n$"? Why is such a map called "one-hot"? Is there a citation for this usage? | |
| Nov 2, 2022 at 8:36 | comment | added | bof | By "not reflexive" do you mean "irreflexive", i.e., that $xRx$ never holds? Are you assuming that for any $x$ and $y$ one and only one of $x=y$, $xRy$, $yRx$ holds? And nothing else is assumed? In other words, is the introduction to your question just an obscure way of describing a tournament of order $n$? Or am I missing something? | |
| Nov 2, 2022 at 8:34 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 2, 2022 at 8:26 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 2, 2022 at 8:23 | comment | added | Arthur B | Thanks for the edits and apologies for the ESL blip, funny as it is. @LSpice I clarified one-hot encoding | |
| Nov 2, 2022 at 6:56 | comment | added | Michael Greinecker | @SamHopkins But general partial orders are not connected. Finite linear orders are certainly interval orders. | |
| Nov 2, 2022 at 1:21 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Nov 2, 2022 at 1:11 | comment | added | Michael Greinecker | The case $d\neq n$ makes no sense. | |
| Nov 2, 2022 at 1:06 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 2, 2022 at 1:05 | comment | added | LSpice | What does "one-hot encoding" mean? (I also took the liberty of changing "monotonous" to "monotonic", which, though less funny, is I think more common.) | |
| Nov 2, 2022 at 0:45 | history | edited | Arthur B | CC BY-SA 4.0 |
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| Nov 2, 2022 at 0:39 | history | asked | Arthur B | CC BY-SA 4.0 |