Timeline for Proving that cosine distance function defined by cosine similarity between two unit vectors does not satisfy triangle inequality
Current License: CC BY-SA 3.0
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| Jan 21, 2017 at 3:14 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 3, 2016 at 20:57 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 2, 2016 at 9:22 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 27, 2016 at 5:26 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 26, 2016 at 22:33 | history | edited | ttnphns |
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| Feb 24, 2016 at 13:30 | comment | added | whuber♦ | @Adam I'm sorry; I never meant to suggest there was anything wrong about your answer. However, I think it could be condensed into a single line in which you show that the $A,B,C$ you chose violate the triangle inequality. I think it would be of greater interest to devote any additional space to explaining how you thought of your counterexample. | |
| Feb 24, 2016 at 13:28 | comment | added | Adam Przedniczek | @whuber OK. My comment has completly wrong overtone. I was only curious if my answer is completly erroneous, because you both with ttnphns were suggesting similar solution. After that I thought my whole understanding of the question was incorrect and it's not so simple. | |
| Feb 24, 2016 at 13:16 | comment | added | whuber♦ | @Adam I didn't say that at all: my suggestions were made only to show how one could think about the cosine similarity and easily arrive at situations that are likely to violate the triangle inequality. Only one such violation--no matter how "extreme"--is needed to show this inequality does not hold. I would also maintain that this line of thinking is substantially simpler than the one you posted, because (1) yours comes without any motivation and (2) there's really nothing left to show once you observe the cosine similarity is a convex function of distance: the conclusion is immediate. | |
| Feb 24, 2016 at 10:25 | comment | added | Adam Przedniczek | @whuber Are you sure that a violation of triangle inequality could only be proven in such an extreme case of almost straight triangles? Could you have a look on a counterexample below, because I think that is much simpler, but after two comments above I'm really confused about it. | |
| Feb 24, 2016 at 1:34 | comment | added | whuber♦ | You do this by finding a counterexample. Since cosine similarity is really a squared distance, look at very small, almost straight triangles for possible violations of the triangle inequality. | |
| Feb 23, 2016 at 13:05 | answer | added | Adam Przedniczek | timeline score: 8 | |
| Feb 23, 2016 at 12:04 | history | edited | Silverfish | CC BY-SA 3.0 |
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| Feb 23, 2016 at 11:59 | review | Close votes | |||
| Feb 23, 2016 at 15:43 | |||||
| Feb 23, 2016 at 11:29 | comment | added | ttnphns | There is a thread stats.stackexchange.com/q/135171/3277 very close to your question. It is about correlation, but correlation is cosine for centered variablers, so it is relevant for your case. | |
| Feb 23, 2016 at 11:21 | review | First posts | |||
| Feb 23, 2016 at 12:05 | |||||
| Feb 23, 2016 at 11:21 | history | asked | Mary | CC BY-SA 3.0 |